A prime number $p$ is ramified in $\mathbb{Q}(\sqrt[p]{a})$.

Using the irreducibility of $X^p-a$ as discussed in the comments write $K=\mathbb{Q}(\sqrt[p]{a})$ and $R=\mathbb{Z}[\sqrt[p]{a}]$. Then $\Delta(R)=\Delta(X^p-a)=(-1)^{\frac{1}{2}p(p-1)}p^pa^{p-1}$, and if we write $a=p^kb$ for $0\leq k<p$ and $(p,b)=1$ then we see that $$ \Delta(\mathbb{Z}[\sqrt[p]{a}])=(-1)^{\frac{1}{2}p(p-1)}p^{p+k(p-1)}b^{p-1}. $$ As $p$ is odd we know that $p+k(p-1)$ is also odd, hence the relation $$ \Delta(\mathbb{Z}[\sqrt[p]{a}])=[\mathcal{O}_K:R]^2\cdot \Delta(\mathcal{O}_K) $$ shows that we must have $p|\Delta(\mathcal{O}_K)$, and hence $p$ ramifies in $K$.

The statement indeed fails for $p=2$. For a quadratic field $\mathbb{Q}(\sqrt{d})$ with $d\equiv 1\bmod 4$ we have $\Delta(\mathcal{O}_K)=d$, which is odd and hence $2$ is unramified in $K$.