Does proof by contradiction assume that math is consistent?

Godel's Incompleteness Theorem does not apply to every mathematical system. However, let us suppose we are working in a system to which it applies. If our system is consistent, your proof of $\neg Q$ is meaningful. If our system is inconsistent, then everything can be proven from it. So your proof tell us that a theorem is a consequence of the axioms of our system.

I think item $4$ in the question introduces an irrelevancy, namely "Rejecting this contradiction". Before you get to item 4, the available information is that $P$ is true and that $Q$ implies not $P$. So $Q$ implies the contradiction "$P$ and not $P$". In both classical and constructive logic, the negation of a statement is equivalent to "that statement implies a contradiction". So we have the negation of $Q$.

The role of the contradiction here is not to frighten us so that we reject it because of our belief in the consistency of mathematics. Rather it is to serve as the (standard) criterion for negation.

If logic is consistent, we have proven Q false.

If logic is inconsistent, then all statements are false (and true, simultaneously).

Either way, Q is false.