I have the following problem written down:
If $\mathcal{L}$ has a finite signature, with no functions, and $T$ is a complete theory with quantifier elimination, then the boolean algebra of $\mathcal{L}$-sentences $\mod T$ is finite.
This seems straightforward, with no functions the only quantifier free sentences are $\mathsf{R c_0 ... c_k}$, and finite signature means there are only finitely many relations and constants to choose from. Of course $T$ might think some of these sentences are equivalent, but that's fine.
Where have I used completeness? Am I wrong or is it unnecessary?
You are correct that completeness is unnecessary. In fact, it is almost certainly an error that completeness was included in the problem statement, since it makes the problem rather trivial: modulo any complete theory, every sentence is either true or false (by completeness!), so the Boolean algebra would have only two elements.