Not sure, please check carefully.
I claim that the array $(h)$ majorates the array $(q)$, that is, $\sum \varphi (h_{ij})\geqslant \sum \varphi(q_{ij})$ for any convex function $\varphi$, in particular for $-\log$, that is your inequality.
Denote the hook lengths of the first (largest) column by $0<c_1<c_2<\dots<c_m$. Then the hooks in $i$-th row, which contains $c_i-i+1$ squares, are all numbers from 1 to $c_i$ except $c_i-c_1$, $c_i-c_2$, $\dots$, $c_i-c_{i-1}$ (this elementary claim is well-known, it shows the equivalence of Frobenius and the hook length formulae.) That is, $$ A:=\sum \varphi(h_{ij})=\sum_i \left(\varphi(1)+\dots+\varphi(c_i)\right)-\sum_{i<j} \varphi(c_j-c_i). $$ Clearly $$ B:=\sum \varphi(q_{ij})=\sum_i \left(\varphi(m-i+1)+\dots+\varphi(c_i+m-2i+1)\right). $$ Take the difference $A-B$ and consider it as a function of $c_m$ only (fix other $c_i$'s). It equals some constant (not depending on $c_m$) plus $$ \sum_{i=1}^{m-1} \varphi(c_m-i+1)-\varphi(c_m-c_i). $$ Each summand increases as a function of $c_m$, since $i-1<c_i$ and $\varphi$ is convex. That is, it is minimal for $c_m=c_{m-1}+1$. Well, so we may suppose that $c_m=c_{m-1}+1$ (two largest rows have equal length). Do this and consider the difference $A-B$ as a function of $c_{m-1}$. Again it is minimal for $c_{m-1}=c_{m-2}+1$, and so on.
If $\varphi$ is strictly convex, then all inequalities are sharp only for a rectangle.