Will the Big Rip tear black holes apart?

The generic thing that happens is that the black hole horizons merge with the cosmological horizon.

To see why this is so, you can consider the case of deSitter Schwarzschild, described in Anon's answer:

$$ dS^2 = f(r) dt^2 - {dr^2 \over f(r)} - r^2 d\Omega^2 $$

$$ f(r) = 1- {2a\over r} - br^2 $$

This has a Nariai limit, described on the Wikipedia page, where the black hole is as large as possible, and this limit is instructive because it has the following properties:

• The black hole horizon and the cosmological horizon are symmetric.
• The space becomes entirely regular, no singularities are present
• The space can deform so that either horizon is the black hole.
• If you deform the space by adding dust between the black hole and the cosmological horizon, you link the space to the Einstein static universe with two black holes present.

The last point is the most relevant, because for two black holes, you know what happens when they meet--- they merge. For a typical observer, which hasn't fallen into a black hole, a deSitter black hole will be attracted to other deSitter black holes and merge. When they become big enough, they are a cosmological horizon, and then the falling of matter into the black hole turns into inflation of the universe into the cosmological horizon smoothly.

If you have a little black hole, and you aren't in it, unless you accelerate, this black hole will merge with the cosmological horizon in your patch. The merger process is an irreversible joining of horizons, analogous to any other such merger. This is the generic situation. To reach the Nariai limit, you need to keep two black holes on opposite sides of the Einstein static universe, and dump all the dust in the universe into them exactly symmetriclally. This is a very unstable process, a small deformation will lead the two black holes to merge into one cosmological horizon, that then eats all the dust.

White hole in expanding de Sitter

$$ds^2 = -dt^2 +\left(dr-\sqrt{k^2 r^2 +2M/r}\;dt\right)^2 +r^2d\Omega^2$$

Black hole in contracting de Sitter

$$ds^2 = -dt^2 +\left(dr+\sqrt{k^2 r^2 +2M/r}\;dt\right)^2 +r^2d\Omega^2$$

Schwarzschild

$$ds^2=-\left( 1-k^2r^2 -\frac{2M}{r} \right)dt^2 +\frac{1}{ 1-k^2r^2 -\frac{2M}{r} }dr^2 +r^2d\Omega^2$$

Believe it or not, they all describe the same metric space in different coordinates.