Assuming P != NP, the answer is surely yes; based on your guarantee, you've restricted the problem to polynomial time machines, and thus the language you describe should be trivially empty. The challenge is proving that a machine that rejects every input truly decides the language--not just finding this machine.
Of course, if P = NP, you have a Rice's theorem problem (and the answer is no). Since the language is now non-trivial, you're describing a non-trivial property of some Turing machines--specifically, the ones bounded by n^c.
The following is drawn from David Richerby's comment above. Given a Turing machine M1, let M2 be a machine that simulates M1 and rejects if it hasn't halted after n^c steps. Deciding whether a machine M1 solves SAT with the promise (that the machine halts after n^c steps) is equivalent to deciding whether M2 solves SAT without the promise. Deciding M2 without the promise is undecidable, by Rice's theorem.