Actually working out a rigorous prediction for Hawking radiation, you need to solve the equations for the relevant QFT semi-classically--treating the metric of spacetime as a substitute for the Minkowski metric usually used in QFT. This is similar to deriving the quantum mechanics of the Hydrogen atom without paying attention to the fact that photons exist and the electromagnetic field is quantized.
Anyway, when you work out the Hawking effect for a scalar field, or a fermion or whatever, what you do is start with the Penrose-Carter diagram for the Schwarzschild spacetime. You define a set of linearized plane wave solutions at past infinity, and then use your curved-space equations of motion for the field to evolve forward at time,and then measure what projection onto a set of linearized plane wave solutions that you get at future infinity. Hawking's result in his original paper was that even if you define the "in" state at past null infinity to be the vacuum state (i.e., the state with no particles), then you STILL have a blackbody distribution at future null infinity--thus, we interpret the black hole as radiating. Note how none of this relies upon virtual pair creation, and how it ties pretty directly into ordinary QFT. The bit about pair creation mostly serves to give a nice physical picture as to why there could be radiation at future null infinity.
Now, there are a few reasons why this procedure will not work for quarks. Marek points out the clearest of them--there is no "in" or "out" state that describes a set of free, unbound quarks or gluons. The best we can do is approximate such a state at high energy. So the semi-classical formalism breaks down on some level.
The second reason that this won't work out is more astrophysical than anything. Black holes are known to likely have a very small physical charge, because if they were to acquire a charge of any size, they would have a very large electromagnetic field that would polarize everything around them, suck in oppositely charged ions, and neutralize themselves. If, by some chance, a black hole did gain some sort of color charge by creating an oppositely colored particle near it, the force on that particle would be so great that it would not be able to escape to infinity, anyway, and the black hole would remain uncolored, and we wouldn't see free quarks escaping to infinity, so the effect you describe shouldn't be possible.
If you're asking whether or not we would see bound states of quarks escaping, however, the answer to that question is almost certainly yes. A pion or a proton is a perfectly good state to do semiclassical analysis with, and (I don't know this for sure, but I'm pretty sure that someone has looked at this before, and there isn't a plausible counterargument that I can think of) that we would get perfectly sensible blackbody radiation out of using them as a test particle, so long as the mass of the black hole is low enough (and therefore, its temperature high enough) to not allow the production of such massive particles.