The answer is **Yes**. This is done in Exercise 120 of my web page link text. You can replace $4$ and $2$ by numbers $n$ and $m$ with $m$ dividing $n$.

**Later**. The required complement. Let me take the situation of the question, but with $A,B,C,D$ being $m\times m$ matrices (thus $n=2m$). If $A$ has an inverse, the Schur complement formula tells you that
$$\det S=\det A\cdot\det(D-CA^{-1}B).$$
Because $A$ and $C$ commute, this is nothing but
$$\det A\cdot\det(A^{-1}(AD-BC))=\det(AD-BC).$$
Now, what if neither $A$ nor $B,C,D$ admit an inverse ? Just do the following. Replace $\mathcal R$ by $\mathcal R(X)$ (rational fractions). Then apply the formula to
$$\Sigma=\begin{pmatrix} A+XI_m & B \\\\ C & D \end{pmatrix},$$
remarking that $A+XI_m$ is non-singular, because its determinant is invertible.

**Even later**. Here are the details when $p:=n/m$ is larger than $2$. The matrix $S$ is $n\times n$, with commuting blocks $A_{ij}\in M_m(\mathcal R)$, $1\le i,j\le p$. To avoid confusion, I use the notation ${\rm Det}$ to denote the determinant in the commutative subring $\mathcal R'$ of $M_m(\mathcal R)$ spanned by the blocks $A_{ij}$, whereas $\det$ is the determinant of an ordinary matrix.

The proof is an induction over $p$.
As in the case $p=2$, I may assume that $A_{11}$ is invertible. Schur's formula gives
$$\det S=\det A_{11}\det S_{11},$$
where
$$ \qquad S_{11}:=\begin{pmatrix} A_{22} & \cdots & A_{2p} \\\\ \vdots & & \vdots \\\\ A_{p2} & \cdots & A_{pp} \end{pmatrix}-\begin{pmatrix} A_{21} \\\\ \vdots \\\\ A_{p1} \end{pmatrix} A_{11}^{-1}\begin{pmatrix} A_{12} & \cdots & A_{1p} \end{pmatrix}.$$
Because $A_{11}$ commutes to every $A_{ij}$, we find
$$\det S ( \det A_{11})^{p-2} = \det T ,$$
where
$$ \qquad T= block(A_{11}A_{ij}-A_{i1}A_{1j})_{2\le i,j\le p}.$$
From the induction hypothesis, and because the blocks $A_{11}A_{ij}-A_{i1}A_{1j}$ commute to each other, one has
$$\det T=\det {\rm Det}((A_{11}A_{ij}-A_{i1}A_{1j}))_{2\le i,j\le p}.$$
Now, in every commutative ring, we have
$$\det((x_{11}x_{ij}-x_{i1}x_{1j}))_{2\le i,j\le p}=x_{11}^{p-2}\det X.$$
Therefore
$$\det T=\det\left(A_{11}^{p-2}{\rm Det} S\right)=(\det A_{11})^{p-2}\det{\rm Det} S.$$
Simplifying by $(\det A_{11})^{p-2}$, we obtain the expected formula
$$\det S=\det{\rm Det} S.$$