Do black holes have zero stress-energy tensor? The Einstein field equations are, in geometrized units:
$$G_{\mu \nu} = 8 \pi T_{\mu \nu}$$
I know that black holes (take the simplest case of a Schwarzschild black hole) are vacuum solutions to the Einstein field equations. Does this imply that $T_{\mu \nu} = 0$? If so, how then does $G_{\mu \nu}$ have nonzero components? If not, what then does $T_{\mu \nu}$ physically represent for a Schwarzschild black hole?
 A: There is more than one usage of the term "vacuum" in physics, which will explain why you may get more than one answer.
First of all, the simplest black hole, called Schwarzschild black hole, is a vacuum solution and it gives for the Einstein tensor
$$
G_{ab} = 0
$$
everywhere that this quantity is well-defined, which is to say everywhere except at the singularity. But note, the Einstein tensor is not the complete information about the spacetime curvature. It is a sum of curvature components in various directions. The Riemann curvature tensor $R^a_{\;bcd}$ is not zero anywhere for this solution (but it tends to zero in the limit $r \rightarrow \infty$, i.e. far from the black hole). The situation is not so different from more familiar physics, where you can have a solution of Laplace's equation $\nabla^2 \phi = 0$ but this does not necessarily imply that $\phi$ itself is zero.
Now to expand a little to other considerations. First, the Schwarzschild black hole has mass (in the sense that other things will orbit it or more generally be attracted to it) and one would like to be able to say where this mass is located. But the stress-energy tensor is zero everywhere ($T_{ab} = 0$) except at the singularity. So it seems we have to say that once any process of collapse which created the black hole has settled down, so that the Schwarzschild metric holds everywhere, then the mass is located at the very place where our theory breaks down! Oh dear. But we can live with this situation as far as practical physics is concerned. In practice, in order to understand how the black hole influences bodies around it, it is enough to say that the mass is located inside (or beyond) the horizon in a spherically symmetric way. 
More fully, one has to see the black hole as a dynamic not a completely static entity, because the metric within the horizon is not static. The goings-on within the horizon are of interest from a theoretical point of view, but have strictly no impact on the rest of spacetime. The mass which influences events outside the horizon is the mass in the past light cone of those events---the mass which was at some stage collapsing before the horizon formed. (This paragraph, and a tweak to the previous one, were added after an exchange of comments with safesphere.)
Finally, a brief comment on Kerr and Reissner-Nordstrom solutions. The former has $T_{ab} = 0$, the latter does not (and all bets are off at the singularity). Therefore, from a GR point of view one would say that the former is a vacuum solution and the latter is not (and I am taking cosmological constant zero throughout). But some people might want to call a region of space with an electric field but nothing else in it a 'vacuum'. That would be quite common terminology for people not interested in the specifically gravitational effects. 
A: This is not a substitute for the nice answers above, but I wanted to add one important point...
The Schwarzschild solution is locally a vacuum solution ($T_{\mu\nu}=0$). But globally it is the solution of a point mass. The point mass is not a part of the spacetime but comes in through boundary conditions on the Weyl tensor.
Globally Schwarzschild is no more a vacuum solution than the voltage field $V=\frac{q}{r}$ is generated by the zero charge density distribution $\rho=0$ on the domain $\mathbb{R}^3-\{r=0\}$. In other words, it is, but there is a point mass/charge at the origin.
A: Black holes are not vaccum solutions to the Einstein field equations. Take for example the Reissner-Nordstrom solution (https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric), or black holes coupled to scalar fields such as the MTZ black hole (https://arxiv.org/abs/hep-th/0406111) or the BBMB black hole(https://www.sciencedirect.com/science/article/abs/pii/0003491674901249). $T_{μν} =0$ (the Schwarzchild case) means that we consider the vaccum case when there is no source for an energy momenum tensor(spacetime is curved only because of the mass of the object).
I cannot understand that part about the Einstein tensor. $G_{μν}$ contains derivatives of the metric function. If we consider the following metric ansatz: 
$$ds^{2} =  -A(r)dt^{2} + B(r)dr^{2} + r^{2}d\theta^{2} + r^{2}sin^{2}\theta d\phi^{2}$$
then: 
$$R_{tt}= -\frac{A'(r)B'(r)}{2B^{2}(r)} + \frac{A''(r)}{2B(r)} + \frac{A'(r)B'(r)}{4B^{2}(r)} - \frac{(A'(r))^{2}}{4A(r)B(r)} + \frac{A'(r)}{rB(r)}=0$$
$$R_{rr}= -\frac{A''(r)}{2A(r)} + \frac{(A'(r))^{2}}{4A^{2}(r)} + \frac{A'(r)B'(r)}{4A(r)B(r)} + \frac{B'(r)}{rB(r)}=0$$
$$R_{θθ}=  -\frac{1}{B(r)} - \frac{rA'(r)}{2A(r)B(r)} +1 + \frac{rB'(r)}{2B^{2}(r)}=0$$ 
since $G_{μν} = R_{μν} - \cfrac{1}{2}g_{μν}R=0\rightarrow R=0 \Rightarrow R_{μν}=0$
Τhe solution to the above equations yields the Schwarzchild solution. Now if we take the solution: $A(r) = 1/B(r) = 1-C/r$ and compute the Einstein tensor of course we will obtain $G_{μν}=0$. 
A: Yes, it implies that $T_{\mu \nu}$ is zero everywhere... except the origin. Just like $G_{\mu \nu}$ is also zero everywhere... except the origin. As it must be, given Einstein's field equations.
Simililarly, the $1/r$ electrostatic potential of a point electric charge is the solution to $\nabla^2 \phi = \kappa \rho$, with $\rho$ zero everywhere... except the origin. Just like $\nabla^2 \phi$ is also zero everywhere… except the origin.  As it must be, given Gauss's law.
(Note: $\phi$ is the electric potential, $\rho$ is the charge density, and $\kappa$ is a constant that depends on your choice of units.)
Just wanted to add this not because the other answers were wrong, but because sometimes people just want the short, simple answer.   
