Sure. Consider something like $T=\mathsf{PA}+\neg\mathit{Con}(\mathsf{ZFC})$ (assuming $\mathsf{ZFC}$ is actually consistent of course).

 - $T$ is consistent but not $\omega$-consistent (it proves that each specific natural number fails to be a $\mathsf{ZFC}$-proof of $\perp$, but still proves that some number is a $\mathsf{ZFC}$-proof of $\perp$).

 - More interestingly, $T\not\vdash\neg \mathit{Con}(T)$. To see this, note that if it did we would have $\mathsf{PA}$-provably that $\neg\mathit{Con}(\mathsf{ZFC})\rightarrow\neg\mathit{Con}(\mathsf{PA})$. Taking the contrapositive and shifting to the stronger theory $\mathsf{ZFC}$, we would get $$\mathsf{ZFC}\vdash\mathit{Con}(\mathsf{PA})\rightarrow\mathit{Con}(\mathsf{ZFC}),$$ a contradiction since $\mathsf{ZFC}$ already proves the consistency of $\mathsf{PA}$.