Is redshift path-dependent? Is redshift dependent simply upon the relative velocity and gravitational potential of transmitter and receiver?
Or is it path dependent; say the receiver runs away for billions of years then changes to the same velocity as the transmitter...
or for any reason the space time in the path becomes locally stretched for a large portion of the path.
And whichever...why?
And what is the evidence either way?
 A: Redshift is path-dependent, but the issue is that in most situations of real physical interest, the photons either arrive through a single path, or the effect is absolutely negligible. 
Let me state some mathematics to explain this. The frequency of a photon with respect to coordinate time $t=x^0$ is proportional to the zeroth component of its wave-vector, we may even set it equal, $k_0 = f$. Now the zeroth component will evolve as
$$ \frac{d k_0}{d t} = -\frac{u^t}{2} g^{\mu\nu}_{\;\;,t} v_\mu v_\nu \,,$$
where $v^\mu =d x^\mu / d t$ and $u^t = dt/d\tau$ (also remember that $v_\mu = g_{\mu\nu} v^\nu$ and $d x^0/d t = dt/dt = 1$). Of course, if $g^{\mu\nu}_{\;\;,t} \approx 0$, the redshift of $k_0$ will depend only on your four-velocity with respect to the static frame; this can be interpreted in terms of a gravitational potential, velocity etc. On the other hand, if the cumulative effect of $g^{\mu\nu}_{\;\;,t}$ cannot be neglected, then talking about gravitational potentials etc. looses meaning.
Now, when $g^{\mu\nu}_{\;\;,t}$ is not negligible, you can be an observer static with respect to $t$ and let an emitter emit two photons of the same frequency in their rest-frame. The first photon goes straight to you, and the second photon goes through a different path which can be arranged either through a gravitational lens or a mirror deflecting it. Generally speaking, the two photons will go through completely different $g^{\mu\nu}$ thus shifting its $k_0$ to completely different values before arriving to you. In other words, generally redshift is very much path-dependent. 
However, our universe is approximately homogeneous and isotropic, and it turns out that in a perfectly homogeneous and isotropic universe any photon, even one sent zig-zagging through space by a system of mirrors, will end up with a shift of $k_0$ that depends only on the ratio of the scale-factors $a$ at the time of its emission and the time of its observation. 
When we include overdensities that actually occur in our universe, this will generally introduce a noise to this perfectly path-independent relation. But if your photon flew at least a few Megaparsecs before getting to you and has not been strongly lensed on the way, you can really trust the redshift to indicate the era the photon comes from. In other words, in cosmology, redshift is mostly path-independent.
As for strong gravitational lensing, this will induce path dependence if the matter causing the lensing is moving with relativistic speeds with respect to the cosmological background, because otherwise $g^{\mu\nu}_{\;\;,t}$ is small (roughly $\sim V/c$ small, where $V$ is the typical velocity in the system). Matter configurations that exhibit such speeds are black-hole and/or neutron-star binaries, as well as various catastrophic events such as supernovas. The redshift of the light that passed through such systems would not be indicative of its original cosmological era. 
However, I believe that observing this phenomenon is highly improbable (and I am quite sure it was never observed). As for experimental confirmation that the redshift was path-dependent: we would have to observe two photons arriving to us from the same source through two paths, and at least one of them passing through such a relativistic, typically extremely variable system! Such double images of the same source occur habitually in strong lensing, but detecting both of them with sufficient accuracy is even more of a lost case then detecting just one image. I.e., I believe there is no experimental or observational evidence of path dependency of redshift, and there will not be any in the near future. 
As for the tests that of the statement that in quasi-stationary fields redshift is path-independent, one should simply refer to the usual experimental tests of gravitational redshift, because a path-dependent theory would necessarily predict deviations from relativistic predictions. On the other hand, I do not believe path-independence for photon paths of cosmological scales can be practically tested, since 1) most double images correspond to photons that do not travel on very different paths on cosmological scales, and 2) we would probably not be able to separate intrinsic variation of redshift within the emitting source (or various other obscuration etc.) from the actual path-dependence of the redshift.
A: Redshift is path-dependent. The photon starts off with some energy-momentum four-vector, and that four-vector gets parallel-transported along the geodesic that is the photon's trajectory.  Parallel transport is path-dependent; that's the definition of curvature.

Is redshift dependent simply upon the relative velocity and gravitational potential of transmitter and receiver ?

Most spacetimes don't have a well-defined gravitational potential. We have a potential only in the case of a static spacetime. For example, cosmological spacetimes aren't static, so there is no potential associated with them. For this reason, it is not meaningful to try to break down cosmological redshifts into kinematic and gravitational parts. We can't even tell whether a distant galaxy is moving relative to us. We can say that it's moving away from us, or we can say equally well that the space in between is expanding.

And what is the evidence either way ?

Interesting question. Gravitational lensing is evidence that the parallel transport of a photon's four-momentum is path-dependent. I don't know of a direct empirical proof that the energy component is path-dependent. This kind of thing is hard to test because I don't think we have any test theory in which it's not true that the energy is path-dependent.
mmeent says:

No, redshift is in general not dependent on the path followed by the light ray. At least not any of the cases we would normally consider (I'll come back to the exception at the end). [...] As far as I can see, the only way that redshift can become path dependent is if the lightray interacts with a gravitational field in such a way that it produces (or absorbs) gravitational waves.

You don't need anything this complicated or special. A much simpler example is the following. In the Schwarzschild spacetime, let a source at rest relative to the gravitating object emit light at event P, and follow two rays that pass symmetrically around the object, reuniting at a point Q directly on the opposite side. An observer at Q will measure unequal redshifts with unit probability if his velocity vector is randomly chosen.
A: The (completely general) formula for red shift in GR is
$$
\frac{\omega_B}{\omega_A} = \frac{ p_\mu(B) v^\mu(B) }{ p_\mu(A) u^\mu(A) }
$$
where $p$ is the 4-momentum of the photon traveling from A to B, $u$ is the  4-velocity of the emitter who observes $\omega_A$ and $v$ is the 4-velocity of the detector who observers $\omega_B$. In this result, $u$ and $v$ have nothing to do with the photon's path (except they are evaluated at A, B), but $p$ has everything to do with the photon's path: it is the tangent 4-vector to the worldline. 
In lensing there is more than one path (null geodesic) between given events A and B. We are used to the fact that different paths arrive at B with different spatial directions; it should not surprise us that they can have different amounts of $p_0$ too, and therefore yes, redshift can depend on path. In order to convince you that this can happen, I'll give a rather impractical example, and leave it to others to find more practical ones.
Suppose that two paths go from A to B, and one of these paths crossed through a local gravitational wave at the phase of that wave where spacetime was locally expanding, but the other path did not. The tidal forces stretch the wavefronts of the lightwave apart from one another. But as the light wave passes out of this gravitational region, the tidal forces fall to zero so do not squeeze it back again. I think such a photon ends up with a different frequency to its 'twin' when they both arrive at B.  
I'm pretty sure this same type of argument can also be made in a static spacetime, but as I say, I'll leave to others to comment on that. 
Such effects are probably too small or too rare to be relevant to observational astrophysics/cosmology, but I guess you never know ....
P.S. on red shift vs. gravitational time dilation
I wrote this answer after reading some comments to the original question in which one person asserts there is no such thing as red shift (of the photon or lightwave itself) because the observed frequency change is wholly owing to a difference in reference frames adopted for the frequency measurements at A and B. This perspective can be useful to training our intuition about GR in some situations. If my high-up caesium clock is ticking faster than your low-down one, then I should not be surprised if microwaves emitted by your caesium atom arrive at me with a rather sluggish oscillation compared to my clock. However, since more generally the frequency change can depend on path, clearly this "it is only about the observers' frames" perspective does not capture the whole truth. So I think there is such a thing as red shift, but it shouldn't be too quickly asserted that the photon itself is changing; rather one should declare all the relevant quantities, including what local clock you are using.
A: 
Is redshift dependent simply upon the relative velocity and gravitational potential of transmitter and receiver?

Neither. Relative velocity is well defined in flat- but not in curved spacetime. @Ben Crowell has already commented on the gravitational potential.
The cosmological redshift $z$ depends on the relative increase of the scale factor $a$ between emission and absorption and is given by $z=(a_{now}-a_{then})/a_{then}$, which can be understood as stretching of the photon's wavelength during the time it travels.
The "stretching" in this sense doesn't mean however that the space is stretched physically. Expansion of the universe doesn't mean more than growing distances between comoving objects. For more see the answer @safespere linked in his comment.     
A: No, redshift is in general not dependent on the path followed by the light ray. At least not any of the cases we would normally consider (I'll come back to the exception at the end).
In almost all cases that we would want to consider (e.g. gravitational lensing, redshift from a source near a black hole, etc.) the metric that we would want to consider is stationary, meaning that the metric has a time translation symmetry or in GR parlence, it has a timelike Killing vector field.
The energy of a lightray is the contraction of this timelike Killing vector with the four-momentum of the lightray. Such contractions is a constant of motion of a geodesic, i.e. it does not change along a geodesic. In particular, it does depend on the exact path taken by the lightray (or equivalently it does not depend on the particulars of the curvature along the objects path).
Of course, generally, the energy of the lightray relative to the timelike Killing vector is not the energy of the light ray in the frame of the emitter, nor that in the frame of the reciever. However, the relation between the three depends only on local quantities at the emitter and reciever (4-velocity of emitter/reciever, the local metric, etc.)
Now, what happens when we let go of the stationarity requirement. The most considered case would that of lightrays in an expanding universe described by an FRWL metric. In that case the metric does not have a time translation symmetry and energy is not necessarily conserved along a geodesic. In fact, the metric does have a preferred timelike vector field defined by the Hubble flow, and the energy of a lightray relative to that timelike vector field does change along the ray. This is the cosmological redshift.
However, by explicit calculation one easily finds that the total gravitational redshift between an emitter and reciever at rest with respect to the Hubble flow is given simply by the ratio of the scale factors at the time of emission and reception. (And for emitters/recievers not at rest with respect to the Hubble flow, the difference is simply the normal relativisitic doppler shift.) In particular, it again does not depend on the details of the path taken or the particulars of how the scale factor has evolved.
As far as I can see, the only way that redshift can become path dependent is if the lightray interacts with a gravitational field in such a way that it produces (or absorbs) gravitational waves. In such a case the amount of energy lost (or gained) certainly depends on the particulars of the chosen path. However, the factional change in energy would be exceptionally tiny in all remotely realistic scenario's I can think of.
