Energies of bodies travelling on different geodesics If two bodies of equal mass are released at rest from point $A$ and travel on different geodesics in a curved spacetime to point $B$ will they have the same total energy at point $B$?
Is the same true of two lightbeams travelling on different null geodesics from point $A$ to point $B$?
 A: It depends on the metric, and what you mean by "total energy."
If the metric is stationary, that is the metric components are independent of time  $\partial_t g_{\mu\nu} = 0$, then the answer is yes.  One way to see this is from the geodesic equation itself.  Normally the geodesic equation is written for the contravariant 4-velocity $u^\mu$ as:
$$
\frac{d}{d\lambda} u^\mu = - \Gamma^\mu_{\alpha \beta} u^\alpha u^\beta \ .
$$
You can also write this in terms of the covariant 4-velocity $u_\mu$ as:
$$
\frac{d}{d\lambda} u_\mu = \frac{1}{2}u^\alpha u^\beta \partial_\mu \ g_{\alpha \beta} \ .
$$
This is a useful form of the geodesic equation because it explicitly tells you about conserved quantities.  For example, if the metric is independent of $t = x^0$ then $\partial_t g_{\alpha \beta} = 0$ and the equation becomes:
$$
\frac{d}{d\lambda}u_0 = 0 \ .
$$
Or, $u_0$ is conserved along the geodesic! If geodesic is that of a massive particle, then $u_0$ is the energy per unit mass.  So, if you release two particles from $A$ with the same mass and $u_0$ in a stationary metric, then at point $B$ they will still have the same $u_0$ and hence the same energy.
It turns out, we can generalize this a bit.  What is important is not exactly that $\partial_t g_{\mu\nu} = 0$, but that the metric (ie. spacetime) has a time symmetry.  Symmetries of metrics are described by Killing Vectors, vector fields which point in the direction of the symmetry.  If $K^\mu$ is a Killing vector of your spacetime, then $K^\mu u_\mu$ will be a conserved quantity along geodesics.  Stationary spacetimes have a Killing vector $(1,0,0,0)$, so $K^\mu u_\mu = u_0$ is conserved.  In general, we say energy is conserved along geodesics if the spacetime has a time-like Killing vector.
Last point: we assumed here your particles were sufficiently light they themselves did not curve spacetime.  Of course that is just an approximation.  If they did curve spacetime, then the metric would be extremely time-dependent, not admit a time-like Killing vector, and energy (ie. $u_0$) would not be conserved along geodesics.  In particular, it would probably be lost to gravitational radiation.
A: If two bodies of equal mass are released at rest from point A and travel on different geodesics in a curved spacetime to point B will they have the same total energy at point B?
Yes. Because gravity is a conservative force. Moreover gravity isn't a force in the Newtonian sense. When two massive bodies fall towards one another, gravity converts potential energy in those bodies into kinetic energy. It doesn't add any energy, conservation of energy applies. So not only do your two minor bodies have the same total energy at point B, their total energy at point B is the same as their total energy at point A. 
Is the same true of two lightbeams travelling on different null geodesics from point A  to point B?
Yes. Gravity doesn't add any energy to a lightbeam. We talk of blueshift, but a descending photon doesn't actually gain any energy. What happens is that when you descend some of your gravitational potential energy is converted into kinetic energy, which you typically radiate away. Then you're left with a mass deficit. Your mass is reduced, your total energy is reduced. So the photon appears to have gained energy when in truth, it hasn't. (It's similar for SR when you accelerate towards a light source and see a blue-shift. The photons haven't changed, you have). Alternatively you can look at this in terms of gravitational time dilation. You and your clocks go slower when you're lower, so the photon frequency looks higher, and since E=hf the energy looks higher too. 
Some people might challenge this, but note that when you send a 511keV photon into a black hole, the black hole mass increases by 511ke/c², not by any other amount. Conservation of energy applies, the books always balance. What isn't conserved is mass. "Invariant" mass varies, see Wikipedia. That's why we have a mass deficit.    
A: The answer is no.
Firstly you talk about points but you should talk about events. An event has a place and a time. It is actually not that easy to come up with events A and B where you can take different geodesics between them, but sometimes it can be done. And your question wouldn't make sense otherwise since if one took longer to get there things could change a lot during the meanwhile.
Secondly you talk about having the same energy. Since they followed different geodesics they started out with different momentum, but let's assume they started out with the same energy.
The answer is no. Firstly GR reduces to Newtonian mechanics in some limits so if we find a Newtonian mechanics problem in the regime where GR and Newtonian gravity agree very well and compute with Newtonian gravity that the energies are very very off then the GR case will have them be different.
So imagine your two very tiny particles take off one going in the y direction at 1mm per billion years and the other one takes off in the x direction at 1mm per billion years then wait until they are 100 trillion light years apart. Then send a sun sized star heading in the y direction with a Jupiter sized gas planet orbiting the sun sized star. We know that it is possible to increase your speed with a slingshot maneuver with a three body problem in Newtonian physics. So have the solar system be heading towards the x moving particle set up just right so that the x moving particle does a slingshot that heads it towards the y moving particle at very high speed and oriented so they intersect at the same time and place.
That's it. You can look at how that sun and planet system curved spacetime and that is an example curved spacetime that had them arrive with radically different energies.
However, there is a sense where you could turn your question into something meaningless and then within that meaninglessness pretend the answer was yes even though it was no. For instance in this set up the y moving particle basically just moved in a straight line for an incredible long time through a region were spacetime wasn't curved much at all (there was about one solar mass about 100 trillion light years away that never came anywhere close to it) and in the natural frame that originally had them moving slowly that's the frame where the x moving one came back with an insane speed.  But energy is totally frame dependant and no frame is better than another. So we could pick a silly frame instead.
So if our particles have the same mass then at event A when they are together we could choose a frame that right near there is the center of momentum frame so they originally had equal and opposite momentum so had equal energy. And then over near event B when they come together again we could again choose a frame that right near there is the center of momentum frame so they end up with equal and opposite momentum so end up with equal energy.  And in GR we can use any frame we want, so we can pick one that near A is that first frame and that near B is that second frame. So regardless of how they move we can measure the frame dependant energy in a way where they are equal.
But that's silly because we can do that no matter how they moved. We could do that if one of the particles found a rocket ship and used that to turn around and come back at high speed and was ejected out of the rocket by the emergency crash system of the rocket. Or if it met a giant cue stick that slammed into it.
