Yes, it is possible. You can add $Con(ZFC)$ as an assumption. If you then derive $\beta$ you have proven within $ZFC$ that $Con(ZFC)\implies \beta$. If you can then prove that $\beta$ is not provable from $ZFC$ you are there.
A trivial example is to take $\beta = Con(ZFC)$. Then ordinary logic proves $Con(ZFC) \implies \beta$ even without $ZFC$.