Can the stress energy tensor have nonzero value in a vacuum region? In general relativity, when solving for the schwartzchild solution, we set $T=0$. 
1) Is it possible for the stress energy tensor to have nonzero value in a vacuum region?
2) Is the stress $T=0$ in a vacuum region in $f(R)$ modified gravity?
 A: No.
But be careful about how you define the word vacuum. For example the Reissner-Nordström black hole has a non-zero stress-energy tensor because the SET includes a contribution from the electrostatic field. So although the vacuum contains no matter it isn't really a vacuum.
A: In the original $f(R)$ gravity, the action is
$$
S = \int \sqrt{-g} \left[ \frac{f(R)}{16 \pi G} + \mathcal{L}_\text{mat}(g^{ab}, \psi) \right] \, d^4x.
$$
Here, $\psi$ represents the collection of matter fields present.  The Euler-Lagrange equation for the metric resulting from this action is
$$
f'(R) R_{ab} - \frac{1}{2} f(R) g_{ab} + \left[ g_{ab} \nabla_c \nabla^c - \nabla_{(a} \nabla_{b)} \right] f'(R) = 8 \pi G T_{ab},
$$
where
$$
T_{ab} = - \frac{\delta \mathcal{L}_\text{mat}}{\delta g^{ab}}.
$$
Now, when we're constructing these models, we usually think of them as "modifying the gravitational dynamics" rather than "modifying the matter dynamics";  so usually we just take the standard matter Lagrangian for whatever matter want to study (electromagnetic fields, perfect fluids, etc.) and drop it into the modified Lagrangian.  So in that sense, $T_{ab} = 0$ in the absence of matter ($\psi = 0$) in $f(R)$ gravity if and only if the same statement holds in conventional gravity.
However, there's one small ambiguity here.  Consider adding a cosmological constant to the above action:
$$
S' = \int \sqrt{-g} \left[ \frac{f(R)}{16 \pi G} + \mathcal{L}_\text{mat}(g^{ab}, \psi) + \Lambda \right] \, d^4x.
$$
Do we view this $\Lambda$ as part of $f(R)$?  Or as part of $\mathcal{L}_\text{mat}$?  It could equally well be viewed as a trivial function of $R$, or as a trivial function of $g^{ab}$ and $\psi$.  But depending on the meaning we assign to $\Lambda$, it will end up contributing to the $f(R) g_{ab}$ term in the first case, or contributing to $T_{ab}$ in the second case.
