The deviations from thermal equilibrium for a black hole are negligible. The outgoing radiation still encodes the exact state. These are not contradictory statements, because a thermal state has a maximum number of microstates which can produce it, so that the microstate doesn't have to leave a macroscopic imprint.

I heard a classical example of this from Sidney Coleman in the early 1990s. Consider heating a lump of coal at absolute zero with a perfectly coherent laser until it glows red hot, emitting photons until it cools down to absolute zero again. The incoming photons are in a pure state, the outgoing photons are thermal. In principle, all the information about the incoming photons are contained in the outgoing photons, but it can be shown that the reduced density matrix of even 50% of all the photons is essentially perfectly random and thermal. You need to look at nearly all the photons to get the density matrix to reveal the substructure.

Any way of experimentally demonstrating the nonthermalness of a black hole will require an impossible measurment of an observable which combines nearly all the photons. The only reasonable way out is to consider a nearly extremal black hole absorbing a tiny amount of matter, and reemitting it. Within string theory, such a process is not thermal at all but is described (in model black holes) by a brane-matter interaction whose output is reasonably calculable. Gubser and others performed this calculation for D-branes scattering gravitons in string theory in 1995, and this method of coherent model black hole incoming/outgoing scattering was one of the major inputs leading to AdS/CFT.

This post imported from StackExchange Physics at 2014-03-17 03:34 (UCT), posted by SE-user Ron Maimon