There are situations where irreducibility is preserved after complexification, however by saying unitary I understand you to mean that the original representation already has an invariant complex structure $J$. In that case the answer is never. The easy way to see this is that the ring of endomorphisms $\text{End}_G(\mathcal{H}_\mathbb{C})$ which commutes with $G$ is not the complex scalars, because it contains independent elements old $J$ (acting on the right factor) and new $i$ (acting on the left) which both square to $-1$. Hence by Schur's lemma, the representation is not irreducible.

If you meant something else by unitary (orthogonal?), then a sufficient condition is $$\text{End}_G(\mathcal{H}_\mathbb{R})=\mathbb{R}$$ I am completely sure about this second part for finite dimensional reps, but it should still hold for Hilbert spaces (I think).

There are situations where irreducibility is preserved after complexification, however by saying unitary I understand you to mean that the original representation already has an invariant complex structure $J$. In that case the answer is never. The easy way to see this is that the ring of endomorphisms $\text{End}_G(\mathcal{H}_\mathbb{C})$ which commutes with $G$ is not the complex scalars, because it contains independent commuting elements old $J$ (acting on the right factor) and new $i$ (acting on the left) which both square to $-1$. Hence by Schur's lemma, the representation is not irreducible.

If you meant something else by unitary (orthogonal?), then a sufficient condition is $$\text{End}_G(\mathcal{H}_\mathbb{R})=\mathbb{R}$$ I am completely sure about this second part for finite dimensional reps, but it should still hold for Hilbert spaces (I think).