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 $\mathsf{PA}$-proving that $\neg\mathit{Con}(\mathsf{ZFC})\rightarrow\neg\mathit{Con}(T)$. Taking the contrapositive and shifting to the stronger theory $\mathsf{ZFC}$, we would get $$\mathsf{ZFC}\vdash\mathit{Con}(T)\rightarrow\mathit{Con}(\mathsf{ZFC}),$$ a contradiction since $\mathsf{ZFC}$ already proves the consistency of $T$. (Why does $\mathsf{ZFC}$ prove the consistency of $T$? If $T$ were inconsistent then $\mathsf{PA}$ would prove $\mathit{Con}(\mathsf{ZFC})$ and hence be inconsistent. So since $\mathsf{ZFC}$ proves that $\mathsf{PA}$ is consistent, $\mathsf{ZFC}$ also proves that $T$ is consistent.)