I'm having a hard time answering this. If the question is really asking whether there is an observer for whom the Kerr metric (rotating BH) reduces to the Schwarzschild metric (BH) in local co-ordinates, then I think the short answer is no. The frame in which the $t\phi$ component of the metric vanishes appears to be counter-rotating with respect to the black hole, so I think it would still see a ring. Basically, it's not the Schwarzschild metric. See this section on Wikipedia.
There are a lot of problems with trying to descibe a stationary/synchronous orbit around a black hole, though. To start with, what is the angular velocity of a black hole? Is it found by considering the angular momentum at the event horizon? Or somewhere else? It might be intuitive to regard the event horizon as the "surface", but it's really a co-ordinate boundary, rather than a real one. Objects that fall in don't see a "surface".
In order to see the ring singularity in a rotating (and electrically neutral) black hole, the observer must be inside the second horizon, where spacetime becomes time-like again. I'd guess a stationary observer in this region would see the ring singularity as some sort of distorted ring. Then again, physics is pretty much breaking down here, so who knows what he'd see.
I'm also not sure whether a synchronous orbit is stable, if it exists. The innermost stable circular orbit is outside the event horizon. I've never seen (nor done myself) a calculation of orbits inside the space connected to the singularity, so I don't know if there are more stable orbits further in. But my instinct is that no stable stationary orbit exists inside the usual innermost one, so circular orbits in the vicinity of the singularity would be unstable too.
Finally, I guess some of this could be circumvented by having a black hole with super-maximal rotation, and therefore a naked singularity.