The black hole does not "evaporate before you reach it", althuogh this is a consistent (but misleading) classical picture. The problem is that any horizon that's been around for a long time has both a white hole and a black hole continuation, and the two look different in terms of horizon properties.
For a white hole horizon, you are classically smooshed on the edge until the explosion, while for a black hole horizon you fall through. In time reverse, for a black hole, you fall through, and stuff coming out is redshifted.
The back-reaction on a black hole is entirely negligible, and only identifiable globally. It's a property of the whole hole, not of one patch of the horizon. The locally Rindler form of a normal black hole shows that you fall through, and any tiny perturbation due to the evaporation is irrelevant.
This is despite the bogus intuition people here seem to have that objects freeze on the horizon and the black hole evaporates under them before they have time to fall in. This intuition is seductive, because it is partly right--- this is the white-hole picture (which is complementary to the black hole picture). But it is false in the sense that the infalling observer is not destroyed at the horizon, but feels nothing special.
The reason one can be sure is because the future (and past) continuations are available when the black hole has been around, and qualitatively insensitive to small perturbations. What you are doing is making a small perturbation and using this as an excuse to switch to a white hole picture, which is not an excuse at all.
The idea that black holes cannot form is the analog of the argument that white holes cannot form by collapse. This doesn't matter, as when the black hole has been sitting there for eons, you can't tell whether it's a black hole or white hole. These things are only made clear once you accept Susskind complementarity.
Mooshing on horizon
The mooshing on the horizon picture is only valid for a proper time that terminates for an infalling observer. The observer is redshifted to oblivion, and merges with the BH horizon (in an exterior picture) after a finite proper time.
But in the observer's local frame, there is nothing singular for an extremely long period of affine parameter as the path becomes null. The argument that a black hole has an interior requires the assumption that when the final explosion is far away, the Hawking radiation behaves semiclassically, it becomes invisible for the infalling observer, so this observer falls through. This is a little bit of a religious point of view in the classical world, because there is no evidence for the interior beyond what you can see in the exterior, but it is justified by the consistency of the quantum picture it gives.
Without knowing that the local equivalence principle holds at the horizon, the argument for Hawing evaporation becomes suspicious. You can use the t-independence of the BH to make what is called a "Boulware vacuum" which is nonradiating, because it conserves the t-notion of energy. This Boulware vacuum
was believed to describe QFT around black holes for a long time. It corresponds to the spacetime around a Schwarzschild black hole which is surrounded by a perfect ideal mirror for everything at (R=2M). This thing is thermodynamically ridiculous in the usual picture, the mirror absorbs thermal energy and doesn't heat up to equilibrium. But this Boulware idea is resurrected every once in a while, in t'Hooft's idea that the black hole has double the correct temperature, for example, because the interior and exterior are identified by a gluing map.
The evidence for the falling-through picture, which is Susskind's, comes most persuasively from the quantum theory. It is this picture that produces AdS/CFT. Without it, it is impossible to understand how black holes become so regular and ordinarily quantum in the extremal limit, where the horizon is still present, but the Hawking radiation goes away.