I'll go with a: In general, no, that's not the case.
Light emission in LED-type devices typically happens when electrons and holes recombine, and the energy that gets freed in that process gets converted to a photon with the resulting wavelength. That happens in the transition zone of a doted semiconductor junction, where there's gradient in the band structure.
Let's imagine a diode in reverse bias: In aforementioned transition zone, there's practically no free charge carriers (no holes and electrons), so the device would be a perfect isolator – I say "would be" if not spontaneous creation of such carrier pairs could happen due to thermal effects (and also, things like photon absorption).
Now, under avalanche breakdown conditions, the electric field across that isolating zone is so high that the charge carriers get accelerated very fast – and might "knock" out other charges from the non-conducting bands (to make this feel a bit more scientific: the electric field gives spontaneously created charges an impulse that is enough to transition further charges in k-space to the conduction band).
Now, these charges will just travel to the contacting areas and recombine there - usually nowhere where there's a) a well-defined bandgap to make emission of visible photons likely and b) no optical structures to couple out that light. You just heat up the substrate.
That is not to say there won't be light emissions in all this: purely from a stochastic point of view, some recombination with visible emissions might happen, and also, nothing says that over the temporal process of that avalanche breakdown, there won't be some times where the whole field configuration wouldn't lead to interesting band diagrams where recombination within the optically relevant parts of the LED might take place, at totally different photon energies than the LED was designed for.