Stand on a black hole Is there an theoretical possibility that an black hole rotates so fast that the acceleration pushing you out of the black hole is nearly equal to the gravity?
 A: There are solutions to Einstein's Field equation (General Relativity) where the ratio of angular momentum to mass is so large that the singularity is visible to the outside instead of being shielded by an event horizon.
But no known astrophysical black hole has a ratio that high. And it looks like when you try to give more angular momentum to an existing black hole that isn't extreme the infalling material has to chase the black hole so quickly to go in the direction that increases the angular momentum that the energy associated with its motion actually increases the mass of the black hole enough to keep that ratio under the extreme critical ratio.
I believe it is still an open question about whether it is always the case that a black hole that is not already extreme can never become that extreme.
A: The quick answer to your question is no, cause nothing can spin that fast.
It's an interesting question, though I think you need to take a step back and ask about formation.    A spinning object will have less gravity around the equator than on the poles, but if an object spins so fast that it has zero gravity, most likely it would fly apart though there was an asteroid discovered that achieved negative gravity around it's rotation, but it holds together by chemical forces.   An object spinning so fast that it has zero gravity in and of itself is quite rare.   
And as objects get smaller, like a skater pulling in their arms, rotation speed tends to increase, but likewise, on a planet or star, so does gravity.    Angular momentum doesn't catch up to gravity, even though Neutron stars can spin very fast, often several rotations a second, gravity still wins.
Jupiter, for example, spins very fast, 1 rotation in 10 hours and it has a visible flattening as a result and measurably less gravity around it's bulging equator than it's flattened poles, both because the poles are closer to the center of mass and because of the angular momentum.
Spinning black holes are rather complicated (Maybe somebody else wants to explain those), but if you have a massively spinning object dense enough to become a black hole, you're still going to get a black hole, at least, around the poles, where gravity is unaffected by the spin and the bulging middle would likely just become an accretion disk, though such an example is unlikely to happen in the space very often cause black holes tend to form inside stars going super or hyper nova, not out of spinning masses like Neutron stars.
I suggest you look up spinning black holes (or Kerr black holes) for some rather interesting reading, but the way you ask the question, it's not possible, a black hole couldn't be de-black holed by spinning it and a spinning mass close to a black hole.   A black hole doesn't have a solid surface so the idea of standing on a spinning black hole isn't very accurate,  you're more accurately talking about orbiting a black hole really fast and the orbital speed equals the speed of light at the photon-sphere outside the black hole, so in theory, a black hole would need to spin faster than the speed of light for you to "fly off of it".
That's probobly a rather clumsy answer that doesn't sufficiently address the relativistic aspects of the question, so take it with a grain of salt.
