Yes. Let $Q$ be the set of sequences that separate a fixed [computably inseparable](https://en.wikipedia.org/wiki/Recursively_inseparable_sets) pair $A$ and $B$, such as the set $A$ of programs that halt with output $0$ and the set $B$ of programs that halt with output $1$. There is no computable separation of $A$ and $B$, and so all elements of $Q$ are noncomputable, but $Q$ is effectively closed, because a finite sequence has no extension in $Q$ just in case it places a $0$ for an element of $A$ or a $1$ for an element of $B$, and this can recognized in finite time.