Do black holes violate T-symmetry?

EDIT: the short answer to this question is that a time-reversed black hole is a white hole, full stop, so if you apply time-reversal to a particle falling into a black hole, you get a particle falling out of a white hole, but we don't physically expect to observe white holes.

Original text:

A blackhole space-time does not violate T-symmetry because, the extended Kruskal solution also contains a white hole:

https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

so, if you time-reverse the portion of a curve falling into the black hole, it will become a portion of a curve falling out of the white hole.

Now, we expect that the universe was created with initial conditions that don't allow white holes to exist, but this would mean that the T-symmetry in GR is spontaneously broken by some quantum theory that is not GR. It is absolutely present in the schwarzschild and Kerr spacetimes, though, thanks to the extend kruskal coordinates trick.

Black holes do not violate T-symmetry, but as macroscopic systems interacting with an environment and subject to the laws of thermodynamics they do have a thermodynamic arrow of time for processes around them.

White hole is simply statistically improbable black hole. Hawking radiation has a thermal spectrum and that means that even high energy and complex states could be radiated away from it. Such energy states would include (for sufficiently large black hole), for example, an astronaut in a spaceship flying out of such a black hole. Of course, the probability of such an event would be immeasurably tiny to occur in our universe, with a realistic black hole much more likely to evaporate by emission of long wavelength photons over the course of $10^{68}$ to $10^{99}$ years (plus the high-energy explosion at the end of it) rather than ever emitting something interesting.

In the (conjectured) distant future of our universe after the baryon matter has decyed there would be a Black Hole Era. At this point it would not even be right to call them black since they would be only things providing illumination (by means of Hawking radiation) to the universe. At this point the thermodynamic arrow of time makes them white holes.

So the really T-symmetric setup would be the black hole at equilibrium with the surrounding space (and so with the same ambient temperature). For such a setup for every photon falling into the black hole there would be (on average) a photon Hawking-radiated away. And for every astronaut Hawking-radiated out of the black hole there would be one manifesting itself as a Boltzmann brain outside of it and then falling in.