Do black holes have a moment of inertia? My question is in the title: Do black holes have a moment of inertia?
I would say that it is: $$I ~\propto~ M R_S^2,$$ where $R_S$ is the Schwarzschild radius, but I cannot find anything in the literature.
 A: Moments of inertia are defined about a given axis of rotation. 

Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation

There exist rotating black holes.

A rotating black hole is a black hole that possesses angular momentum. In particular, it rotates about one of its axes of symmetry.

Thus by definition a black hole, since it is massive , must have a moment of inertia, for details see this link, or this link.
A: The angular velocity of a Kerr black hole with mass $M$ and angular momentum $J$ is
$$ \Omega = \frac{J/M}{2M^2 +  2M \sqrt{M^2 - J^2/M^2}} $$
The moment of inertia of an object can be thought of as a map from the object's angular velocity to its angular momentum. However, here we see that the relationship between these two quantities is non-linear. If we want to think of moment of inertia in the usual sense, we should linearise the above equation. When we do so, we find the relationship
$$ J = 4 M^3 \Omega \qquad (\mathrm{to\ first\ order})$$
And so the moment of inertia is
$$ I = 4 M^3 $$
In other words, the expression you guessed is correct, and the constant of proportion is unity. Note that since the Schwarschild radius of a black hole is merely twice its mass, and since the only two parameters that describe the black hole are its mass and angular momentum, any linear relationship between the angular velocity and angular momentum of our black hole must be of the form $J = k\, M R_S^2\, \Omega$ on dimensional grounds.
Note that $G = c = 1$ throughout.

EDIT. 
As pointed out in the comments, it's not obvious how one should define the angular velocity of a black hole. At the risk of being overly technical, we can do this as follows. First consider the Killing vector field $\xi = \partial_t + \Omega \partial_\phi$ (using Boyer-Lindquist coordinates), where $\Omega$ is defined to be as above. The orbits, or integral curves, of this vector field are the lines $\phi = \Omega t + \mathrm{const.}$, which correspond to rotation at angular velocity $\Omega$ with respect to a stationary observer at infinity.
One can show that this vector field is tangent to the event horizon, and its orbits lying on the event horizon are geodesics. These geodesics hence rotate at angular velocity $\Omega$ (with respect to an observer at infinity), and hence it is natural to interpret the quantity $\Omega$ as the angular velocity of the black hole. Whether it is possible to make a more definite statement than this I do not know.
