Procedures 1 and 2 are equivalent when the action is quadratic in the momentum, and when there is a gauge fixing which produces a unitary quantum field theory. Unitarity is not obvious in the Path integral, as immediately noted by Dirac. It is established either by proving reflection positivity in the path integral directly, or by passing to a canonical description where the unitarity is obvious, because the Hamiltonian is real.
It is important to note that the fact that the quantities in the path integral are not operators is completely insignificant. Their products don't commute, and require careful definition in terms of time order, which is in every way corresponds to the ordering ambiguities in the Hamiltonian formalism. If you want to think of them as operators, you can, they act on the boundary conditions coming in in the same way as Heisenberg operators, because they are just the matrix elements of Heisenberg operators. There is no difference in the properties of the quantities in the path integral formalism and any other formalism, they don't get easier in the path integral.
Feynman-Fourier transform
For Hamiltonians which are not quadratic in the momentum, it is trickier to pass to a path integral, the quantum action is not equal to the classical action. The general prescription to pass to the Feynman description is through the phase-space path integral:
$$ K(x,y) = \int Dx Dp e^{i\int (p\dot{x} - H(p,q))} dt$$
where the term $p\dot{x}$ is to be interpreted as $p_t(x_{t+\epsilon} - x_t)$, that is, $\dot{x}$ is a forward difference, and H(p,q) is "normal ordered", meaning all p terms are commuted to appear first.
Then the Feynman form is given by integrating out the momentum. This cannot be done in closed form in general, so there are many examples of well-behaved Hamiltonians whose Lagrangian description is not closed-form expressible, for example
$$H= p^4 + V(x)$$
and there are converse examples of nice Lagrangians whose Hamiltonian form is not very nice. I will give such an example in a minute. But first, the Feynman transform.
When the Hamiltonian is of the form
$$H =K(p) + V(x)$$
Then the Lagrangian description is expressed entirely in terms of the function K' appearing in this formula:
$$ e^{-K'(v)} = \int e^{-K(p) + i p v} dx$$
That is, the Feynman transform K' is the log of the fourier transform of the exponential of minus the original function. To see that this works is simple, you Wick rotate each integral over p and do the integral formally using K'.
Each exactly expressible Feynman transform is interesting, but there are very few. In the literature, there is exactly one:
$$ K(p) = {1\over 2} p^2 \implies K'(v) = {1\over 2} v^2 $$
This takes care of quadratic momentum. If you restrict your attention to the published literature, the Feynman integral table is that ridiculous. This much takes care of all the usual quantum field theories, however, so it is not insignificant.
More Feynman transforms
Since the literature on this is pathetic, here are some nontrivial Feynman transforms, and the physics they describe:
Cauchy quantum mechanics: $$ K(p) = |p| \implies K'(v) = -\log( 1+ x^2) $$
This is a nice transform, because the Lagrangian path integral you get (in imaginary time) is
$$\int Dx e^{-\int \log(1+|\dot{q}|^2) - V(x)}$$
This path integral defines a path integral over Levy flights whose stable distribution is the Cauchy distribution. You can see this by looking at the propagation function between adjacent times, it gives a Cauchy distribution. This path integral defines Cauchy quantum mechanics. It's a special case of
Levy quantum mechanics: $$ K(p) = |p|^\alpha \implies K(x) = - \log( L_\alpha(x) ) $$
For $0<\alpha<2$, and where $L_\alpha$ is the unit Levy stable distribution for exponent $\alpha$. These quantum mechanical systems have been studied in recent years, but their path integral doesn't appear anywhere in the literature. The path integral is given by the Feynman transform.
There are tons more interesting Feynman transforms, they are the analog of Legendre transforms in classical mechanics, and are just as useful.
Proving unitarity
The path integral is well defined for any Euclidean statistical theory, but only a very few of these continue to quantum mechanics. A proof of unitarity usually passes to a Hamiltonian formulation, because this is manifestly unitary.
An example of a nonunitary renormalizable path integral statistical system which is otherwise perfectly ok is
$$\int d^8x |\nabla \phi|^4 + Z|\nabla\phi|^2 + t(\phi)^2 + \lambda \phi^4 $$
This system was studied in $8-\epsilon$ dimensions by Mukhamel, because it's epsilon expansion is pretty much the same as the $4-\epsilon$ expansion of the $\phi^4$ model. In eight dimensions, it defines a perfectly good second order point when Z and t are tuned to the right values. But the theory is absolutely not unitary--- there aren't any interacting scalar quantum theories in 8 dimensions. This can be seen immediately from the Kallen representation.
Any propagator in a unitary theory can be expressed in Euclidean space as
$$ G(k) = \int ds {\rho(s) \over k^2 - s}$$
That is, as a superposition of ordinary propagators at different values $s$ of the squared mass. $\rho(s)$ is non-negative, because in real time it is the norm of the state created by the field whose propagator you are expressing. It is this representation that tells you that wrong sign poles are ghost states.
The ${1\over k^4 + (A+B) k^2 + AB}$ Mukhamel Lifschitz-point propagator (with strange parametrization) is expressible as a spectral representation by partial fractions:
$$G(k) \propto {1\over k^2 + A} - {1\over k^2 + B}$$
This Kallen-Lehman spectral representation is clearly ghosty. The double-pole case has a Lehman function which is the derivative of a delta function, which is not positive definite either, by limits.
There are tons of non-unitary Euclidean theories, and to find the unitary ones, the Hamiltonian formulation is very helpful. Finding a no-ghost gauge and transforming to canonical form is how you prove that Gauge theory is unitary, for example.