Black holes are solutions of the Einstein equations, the fundamental equations of Einstein's relativity, that link together space-curvature to density of matter and energy. The black holes solutions are found in the void, when you let a single point of the space-time be singular (the 'center of the black hole'). If you further assume that the black hole is static, you find three solutions:
- the Schwarzschild solution, this is a non-rotating non-charged black hole, which is entirely described by its mass M ;
- the Kerr solution, this is a rotating non-charged black hole, entirely described by its mass and spin (how fast it rotates) ;
- the Kerr-Newmann solution, describing a rotating charged black hole, entirely described by its mass, spin and charge.
It is a reasonable interpretation to think about a black hole as being a void with an infinitely small singularity in which lays all the mass, momentum and charge. In this sense, a black hole is empty, except for the region not described by the space-time framework (and that is not subject to the Einstein equations). So it is correct to think that black holes are only distorted region of space-time. The space-time is distorted so that going from the infinity (where there is no mass and no spin) you can reach the singularity continuously. If you fall onto a rotating black-hole, you will end up with the same spin as the black hole once you hit the singularity, (almost) regardless of how much spin you had initially.
Schwarzschild black holes force you to fall inward once you cross the 'horizon' of the black hole, regardless of how much acceleration you can produce. Kerr black holes have the same horizon, as well as a so-called 'ergosphere'. This 'sphere' is a region of space where everything including light is forced to rotate in the direction of the spin of the black-hole.
In the case of the Earth, you cannot directly use the Kerr solution to approximate the rotating Earth (as stated previously, this is a solution of the equations in void) how much drag you will have, but this gives you the correct idea, as stated by @James K.