Let $$M = \left[\begin{matrix} M_{1,1} & M_{1,2} & \cdots & M_{1,n}\\ M_{2,1} & M_{2,2} & \cdots & M_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ M_{n,1} & M_{n,2} & \cdots & M_{n,n} \end{matrix}\right] \in \mathbb{F}_q^{nm \times nm}$$ where the $M_{i,j} \in \mathbb{F}_q^{m \times m}$ are invertible for all $i, j \in [n]$.
Consider the partial transpose $M^{\text{BT}}$: $$M^{\text{BT}} = \left[\begin{matrix} M_{1,1} & M_{2,1} & \cdots & M_{n,1}\\ M_{1,2} & M_{2,2} & \cdots & M_{n,2}\\ \vdots & \vdots & \ddots & \vdots\\ M_{1,n} & M_{2,n} & \cdots & M_{n,n} \end{matrix}\right]$$ wherein the blocks' positions are transposed but the blocks themselves are not transposed.
Is it true that $\det(M) \neq 0$ if and only if $\det(M^{\text{BT}}) \neq 0$?
The answer to your question is No.
A theorem of Silvester,
gives for $2\times 2$ block matrix $$ M=\begin{bmatrix}A & B\\ C & D \end{bmatrix} $$ with $CD=DC$, we have $$ \det M= \det (AD-BC). $$
We try finding a counterexample in $2\times 2$ block matrices consisting of $2\times 2$ matrices.
Let $$ A=\begin{bmatrix} 1& 1 \\0&1 \end{bmatrix}, \ B=\begin{bmatrix} 7 &-1\\-8&4\end{bmatrix}, \ C=\begin{bmatrix} 0&-1\\1&0\end{bmatrix}, D=\begin{bmatrix} 1&-1\\ 1&1 \end{bmatrix} $$
We have $CD=DC$. Then by Silvester's theorem, $$ \det M=\det(AD-BC)=\det \begin{bmatrix} 3 &7 \\-3&-7\end{bmatrix}=0. $$
However, $$ M^{BT}=\begin{bmatrix}A & C\\ B & D \end{bmatrix} $$ has $$ \det M^{BT} = 3. $$
Thus, by taking sufficiently large $q$, for example, $q=31$, we see that your question is answered negatively.