If ZFC is consistent, then NBG does not prove the second-order $\in$-recursion scheme. 

To see this, take an $\omega$-nonstandard model of NBG, with only the parametrically definable classes. For each standard $n$, there is a class $\Sigma_n$ truth predicate. For a class to be a $\Sigma_n$-truth predicate is a first-order expressible property about that class, since one need only assert that it fulfills the Tarski recursion. Meanwhile, there can be no definable truth predicate for nonstandard $\Sigma_n$ truth, by the usual proof of Tarski's theorem. 

So there can be no least $n$ for which there is a $\Sigma_n$-truth predicate, and this violates second-order $\in$-recursion.

We can turn this argument into a proof that the second-order $\in$-recursion scheme implies that there is a truth predicate for first-order truth. There is such a predicate for $\Sigma_0$ truth, and from any such $\Sigma_n$ truth predicate we can define a $\Sigma_{n+1}$-truth predicate. These are unique when they exist and cohere into a full truth predicate for first-order truth. 

This implies $\text{Con}(\text{ZFC})$ and much more, since indeed it implies that the universe $V$ is the union of an elementary chain of $V_\alpha$s, each of which will be a transitive model of ZFC.