To rephrase the question slightly, you are asking for one of the Betti numbers of the (3+1)-dimensional manifold corresponding to one of the solutions of the Einstein field equations that corresponds to charged or rotating black hole.
The Betti numbers of a manifold are topological invariants that intuitively represent the number of non-contractible d-dimensional "handles" on that manifold (or formally, the number of generators of the d-th homology group), and for a D dimensional manifold there are D+1 nonzero Betti numbers, $B_0,B_1, ... ,B_D$. The zero-th, $B_0$, corresponds to the number of connected components, $B_1$ the number of one-dimensional handles, and so on.
For example, the Betti numbers of a single torus $\mathbb{T}^2$ are: $B_0=1$, $B_1=2$, and $B_2=1$; and for a sphere $\mathbb{S}^2$ they are $B_0=1$, $B_1=0$, and $B_2=1$. These can be related directly back to simpler invariants you may already have heard of, like the Euler characteristic $\chi$, and the genus $g$, and we can see that for closed two-dimensional manifolds with no boundaries like the above, $g$ is related to the first Betti number:
$B_1 = 2g$
For surfaces where $B_1$ is not an even number, that is, for surfaces which are open or aren't orientable, this definition breaks down, and we talk instead of a non-orientable genus $k=B_1$. For higher dimensional manifolds we can take $k=B_{D-1}$.
So to return to your question, we want to know what the Betti numbers of the spacetime you're describing are. Assuming the topology of timelike curves is simple enough (i.e. no closed timelike curves) or that the solution is independent of time, it's easiest to look at the topology of the purely spacelike part of the solution; $D=3$, so $k=B_2$.
@Siva pointed out that the only relevant non-contractible surface that is introduced by the event horizon is one sphere, $\mathbb{S}^2$, so I would guess* that $B_2 = 1$. This would mean that $k=1$, which is a genus, but because $B_2$ is not an even number, essentially because asymptotically flat spacetimes like these are open, not closed, we can't interpret it in the same sense as the genus of a torus or another 2-d handlebody. But it is a genus in some modified sense, it's a non-orientable genus.
So, I would say that $k=1$.
*If there is a string theorist or someone else who can compute homology groups better than me, this might need checking. In 2D there's an isomorphism between the first homotopy group (which is really what Siva's sphere argument is pointing to) and the first homology group but this doesn't necessarily hold in higher dimensions IIRC. I did a quick calculation with a cellular homology that looked reasonable but I might be oversimplifying.
