I have a non-Lagrangian group $G$ of order $pq^3$, $Q$ a Sylow $q$-subgroup of G and a $H$ a subgroup of $Q$ with $|H|=q^2$. It is clear that $Q \subseteq N_G(H)$. I must prove that $G$ doesn't posses subgroups of order $pq^2$.
I supposed that $G$ posses a subgroup $P$ of order $pq^2$. $P$ will posses a normal Sylow $p$-subgroup or a normal Sylow $q$-subgroup. In the first case, $P$ will contain a subgroup of order $pq$ , making $G$ a Lagrangian group, a contradiction. But I don't find a way for the second case.
Suppose $P \le G$ with $|P|=pq^2$, $|G|=pq^3$. As you say, if $P$ has a normal Sylow $p$-subgroup, then $G$ has subgroups of all possible orders.
So suppose that $P$ has a normal Sylow $q$-subgroup $H$. Since $H$ is normal in both in $P$ and in a Sylow $q$-subgroup of $G$, we must have $H \lhd G$. Let $R$ be a Sylow $p$-subgroup of $P$. If $|N_P(R)|>p$, then $N_P(R)$ contains a subgroup of order $pq$.
Otherwise, by the Frattini argument, we have $|N_G(R)|=q|N_P(R)|=pq$.