Let $G$ be a finite group and $H$ a normal subgroup of $G$. Let $f:G\rightarrow G/H$ be the canonical homomorphism.
Let $Q\leq G/H$ be a $p$-subgroup of $G/H$. I have to show that $f^{-1}(Q)\leq G$ is a $p$-subgroup of $G$.
Since $f$ is surjective, we have that $H\trianglelefteq f^{-1}(Q)$ and $f^{-1}(Q)/H\simeq Q$. The latter implies that $[f^{-1}(Q):H]=|Q|$. I also know that $$|f^{-1}(Q)|=[f^{-1}(Q):H]\cdot|H|;$$ thus $|f^{-1}(Q)|=|Q|\cdot|H|$. I am not sure where to go from here.
Edit:
Is this even possible without additional assumptions? This would require $H$ itself to be a $p$-group, right?
As stated, the result is false. Take $G=S_3$, $H$ the nontrivial normal subgroup of $G$. Then $G/H$ is cyclic of order $2$, hence a $2$-group. Taking $Q=G/H$ we have that $f^{-1}(Q)=G$ has order $6$, not a $2$-group.
The stated result holds if and only if $H$ is itself a $p$-group (same $p$). The proof that this holds is exactly what you give: since $|H|$ divides $|f^{-1}(Q)|$, if there is a prime $q$ that divides $|H|$ with $q\neq p$, then $f^{-1}(Q)$ contains an element of order $q$ (by Cauchy's Theorem) and hence is not a $p$-group. Conversely, if $H$ is a $p$-group, then $|f^{-1}(Q)|$ is a product of two powers of $p$, hence itself a power of $p$.