Is the total energy of the universe zero?
On my blog, I published a popular text why energy conservation becomes trivial (or is violated) in general relativity (GR).
To summarize four of the points:
In GR, spacetime is dynamical, so in general, it is not time-translation invariant. One therefore can't apply Noether's theorem to argue that energy is conserved.
One can see this in detail in cosmology: the energy carried by radiation decreases as the universe expands since every photon's wavelength increases. The cosmological constant has a constant energy density while the volume increases, so the total energy carried by the cosmological constant (dark energy), on the contrary, grows. The latter increase is the reason why the mass of the universe is large - during inflation, the total energy grew exponentially for 60+ $e$-foldings, before it was converted to matter that gave rise to early galaxies.
If one defines the stress-energy tensor as the variation of the Lagrangian with respect to the metric tensor, which is okay for non-gravitating field theories, one gets zero in GR because the metric tensor is dynamical and the variation — like all variations — has to vanish because this is what defines the equations of motion.
In translationally invariant spaces such as Minkowski space, the total energy is conserved again because Noether's theorem may be revived; however, one can't "canonically" write this energy as the integral of energy density over the space; more precisely, any choice to distribute the total energy "locally" will depend on the chosen coordinate system.
(Now I notice you're the same person who asked this at MathOverflow, where I've previously answered something similar -- if you didn't like the answer then, you won't like it now.)
This is really just expanding on Marek's comment:
How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.
What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the whole action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.
This is what "canceled out by the negative energy of the gravitational field" means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.
The claim that the total energy in the universe is zero can be rigorously justified.
To answer your specific questions:
General Relativity is required. It does not apply for Newtonian gravity.
It has to be assumed that classical general relativity, with or without cosmological constant, is correct and that the universe is spatially homogeneous on sufficiently large scales. If the universe is infinite the total energy is not really defined, but it is still true that the total energy in an expanding volume of space is asymptotically zero when the region is large enough for the homogeneity of the universe to be a good enough approximation.
Here is a link to a paper as requested.