This is an old, hard, controversial question. It is in some sense not well defined, because there are subtle ways in which it can be difficult to pin down the distinction between a radiation field and a non-radiative field. Perhaps equivalently, there are ambiguities in the definition of "local." If an accelerating charge did radiate, it would cause a problem for the equivalence principle.
There are arguments by smart people who claim that an accelerating charge doesn't radiate (Harpaz 1999; Feynman's point of view is presented at http://www.mathpages.com/home/kmath528/kmath528.htm ). There are arguments by smart people who claim that an accelerating charge does radiate (Parrott 1993). There are other people who are so smart that they don't try to give a yes/no answer (Morette-DeWitt 1964, Gralla 2009, Grøn 2008). People have written entire books on the subject (Lyle 2008).
A fairly elementary argument for the Feynman point of view is as follows. Consider a rigid blob of charge oscillating (maybe not sinusoidally) on the end of a shaft. If the oscillations are not too violent, then in the characteristic time it takes light to traverse the blob, all motion is slow compared to c, and we can approximate the retarded potentials by using Taylor series (Landau 1962, or Poisson 1999). This procedure will lead us to compute a force and therefore the lower derivatives (x'') from the higher derivatives (x'''); but this is the opposite of how the laws of nature normally work in physics. Even terms in the Taylor series are the same for retarded and advanced fields, so they don't contribute to radiation and can be ignored. In odd terms, x' obviously can't contribute, because that would violate Lorentz invariance; therefore the first odd term that can contribute is x'''. Based on units, the force must be a unitless constant times $kq^2x'''/c^3$; the unitless constant turns out to be 2/3; this is the Lorentz-Dirac equation, $F=(2/3)kq^2x'''/c^3$. The radiated power is then of the form $x'x'''$. This is nice because it vanishes for constant acceleration, which is consistent with the equivalence principle. It's not so nice because you get nasty behavior such as exponential runaway solutions for free particles, and causality violation in which particles start accelerating before a force is applied.
Integration by parts lets you reexpress the radiated energy as the integral of $x''x''$, plus a term that vanishes over one full cycle of periodic motion. This gives the Larmor formula $P=(2/3)kq^2a^2/c^3$, which superficially seems to violate the equivalence principle.
Note that starting from the expression $x'x'''$ for the radiated power, you can integrate by parts and get $x''x''$ plus surface terms. On the other hand, if you believe that $x''x''$ is more fundamental, you can integrate by parts and get $x'x'''$ plus surface terms. So this fails to resolve the issue. The surface terms only vanish for periodic motion.
In a comment, Michael Brown asks the natural question of whether the issue can be resolved by experiment. I don't know that experiments can resolve the issue, since the issue is really definitional: what constitutes radiation, and how do we describe the observer-dependence of what constitutes radiation? In particular, if observers A and B are accelerated relative to one another, it's not obvious that what A calls a radiation field will also be a radiation field according to B. We know that bremsstrahlung exists and that it's the process responsible for the x-rays that produce an image of my broken arm. There doesn't seem to be much controversy over whether the power generated by the x-ray tube can be calculated according to $x''x''$. What about the frame of the decelerating electron, in which $x''=0$? The question then arises as to whether this frame can be extended far enough to encompass the photographic film or CCD chip that forms the image.
It gets even tougher when we deal with gravitational accelerations. To a relativist, a charge sitting on a tabletop has a proper acceleration of 9.8 m/s2. Does this charge radiate? How about a charge orbiting the earth (Chiao 2006) or free-falling near the earth's surface? Lyle 2008 has this clear-as-mud summary (gotta love amazon's Look Inside! feature):
To a first approximation, remaining close enough to the charge for curvature effects to be negligible, in the sense that the metric components remain roughly constant, GR+SEP tells us that there should not be electrogravitic bremsstrahlung for a charge following a geodesic, although there will when the charge follows curves [satisfying the equations of motion], due to its deviation from the geodesic.
Unfortunately, calculations show that the electromagnetic radiation from a free-falling charge, if it exists as suggested by the Larmor $x''x''$ formula, would be many, many orders of magnitude too small to measure.
Chiao, http://arxiv.org/abs/quant-ph/0601193v7
Gralla, http://arxiv.org/abs/0905.2391
Grøn, http://arxiv.org/abs/0806.0464
Harpaz, http://arxiv.org/abs/physics/9910019
Landau and Lifshitz, The classical theory of fields
Lyle, "Uniformly Accelerating Charged Particles: A Threat to the Equivalence Principle," http://www.amazon.com/Uniformly-Accelerating-Charged-Particles-Equivalence/dp/3540684697/ref=sr_1_1?ie=UTF8&qid=1373683154&sr=8-1&keywords=Uniformly+Accelerating+Charged+Particles%3A+A+Threat+to+the+Equivalence+Principle
C. Morette-DeWitt and B.S. DeWitt, "Falling Charges," Physics, 1,3-20 (1964); copy available at https://journals.aps.org/ppf/abstract/10.1103/PhysicsPhysiqueFizika.1.3
Parrott, http://arxiv.org/abs/gr-qc/9303025
Poisson, http://arxiv.org/abs/gr-qc/9912045
Best Answer
Yes, it will radiate. The light does not come from inside the black hole. The accelerated charge of the black hole affects the electromagnetic field around but outside the event horizon.
You could ask the same question you are asking now about gravitational waves. Binary black hole systems radiate gravitational waves, even though a gravitational wave that fell into a black hole could not escape.