Complexity of black box satisfiablty
If it's a black box, you can't reduce it to SAT.  That's exactly what it means to be a black box: a black box is one where you can feed in any input $x$ you like, and get back $f(x)$, but where you know nothing about the structure of $f$ (for instance, you are not given an algorithm for $f$, you do not know the CNF for $f$, and so forth).

In particular, given only black-box access to $f$, in the worst case, you need $2^n$ queries to the black box to determine whether $f$ is satisfiable: you need to evaluate $f$ on every possible input, in the worst case.  Consider any algorithm which makes at most $2^n-1$ queries to the black box.  Suppose it got back $0$ as the answer to every query.  Then there is some input, say $x'$, which has not been queried.  Now the correct answer as to whether $f$ is satisfiable depends upon whether $f(x')=0$ or $f(x')=1$, but the algorithm has no clue which is the case.  No matter what the algorithm outputs, an adversary could always pick a $f$ (after the fact) that is consistent with all of the queries made to the black box but that makes the algorithm's output wrong.  Therefore, no deterministic algorithm can do better than $2^n$ queries, in the worst case.

It's not in EXPTIME, and it's not in NP, because it is an oracle problem, whereas NP and EXPTIME are for non-oracle problems -- but it is in the analogue of EXPTIME for oracle problems.... and in the analogue of NP for oracle problems, too.