It's true for $d = 2$. Even for $d = 3$, it fails miserably. If $K = \mathbf{Q}(p^{1/3})$ and $p \ne 3$ then $p^2$ exactly divides $\Delta_{K}$, whereas $p^4$ exactly divides the discriminant of the Galois closure. (As David points out, things are even worse for $p = 3$.) Not to mention the fact that $p$ doesn't divide to any power the discriminant of $\mathbf{Q}(\sqrt{-3})$ which is contained inside the Galois closure of $K$.


 About the only time the power of $p$ dividing the discriminant of the Galois closure is $n!/2$ times the power of $p$ dividing the discriminant is when $p$ _exactly_ divides $\Delta_K$. In this case, the power of $p$ dividing the fixed field of $H$ in $S_n$ is equal to


$$(n-2)! \cdot \frac{\text{the number of $2$-cycles which do not lie in $H$}}{|H|}$$

This is always an integer, and it answers your question when $\Delta_K$ is squarefree. (The general case having already seen to be false.)