I will write $R$ for $R_{q}$, because $q$ is constant. In the following, I am going to assume that $R$ is an arbitrary right inverse of $M_{q}$, rather than the specific right inverse $M_{q}^{T}\left( M_{q}M_{q}^{T}\right) ^{-1}$ which you have suggested. This makes the statement a tad more general and rids us of a red herring.
I shall prove that for any $i\in\left\{ 1,2,...,n\right\} $, $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, the value $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$ (independently, and independent, of the choice of right inverse $R$).
For every $\ell\in\left\{ 1,2,...,n\right\} $, we abbreviate $C_{q\ell}$ by $v_{q}$. Thus,
$v_{\ell}=C_{q\ell}=\left( -1\right) ^{q+\ell}\det\left( \underbrace{M\text{ without row }q}_{=M_{q}}\text{ and column }\ell\right) $
(1) $=\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column }\ell\right) $.
From this it easy to obtain
(2) $\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}=0$.
[Proof of (2): Let $G$ be the result of inserting a copy of row $j$ of the matrix $M_{q}$ between the rows $q-1$ and $q$ of this matrix. Then, $G$ is an $n\times n$-matrix with two equal rows, and hence has determinant $\det G=0$. But Laplace expansion of $\det G$ along the row we have inserted yields
$\det G=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column }\ell\right) $,
since the entries of this row are $\left( M_{q}\right) _{j\ell}$ and the cofactors corresponding to this row are $\left( -1\right) ^{q+\ell} \det\left( M_{q}\text{ without column }\ell\right) $. Now, (1) yields
$\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }v_{\ell}$
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column } \ell\right) $
$=\det G=0$,
so that (2) is proven.]
Since $M$ is generic, its cofactors $v_{1}$, $v_{2}$, $...$, $v_{n}$ are nonzero.
Fix $i\in\left\{ 1,2,...,n\right\} $. Let $N_{i}$ denote the matrix $M_{q}$ without column $i$. Thus, $\det\left( N_{i}\right) =\det\left( M_{q}\text{ without column }i\right) =\left( -1\right) ^{q+i}v_{i}$ (because (1) yields $v_{i}=\left( -1\right) ^{q+i}\det\left( M_{q}\text{ without column }i\right) $).
Let $S_{i}$ denote the matrix $\left( R_{jk}-\dfrac{v_{j}}{v_{i}} R_{ik}\right) _{j\in\left\{ 1,2,...,n\right\} ;\ k\in\left\{ 1,2,...,n-1\right\} }$. The $i$-th row of this matrix $S_{i}$ is zero; let $S_{i}^{\prime}$ denote the matrix $S_{i}$ without row $i$.
Now, we claim that $N_{i}S_{i}^{\prime}=I_{n-1}$ (the $\left( n-1\right) \times\left( n-1\right) $ identity matrix). Indeed, for every $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$, the $\left( j,k\right) $-th entry of $N_{i}S_{i}^{\prime}$ is
$\left( N_{i}S_{i}^{\prime}\right) _{jk}=\sum_{u\in\left\{ 1,2,...,n-1\right\} }\left( N_{i}\right) _{ju}\left( S_{i}^{\prime }\right) _{uk}$
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} }\left( M_{q}\right) _{j\ell}\underbrace{\left( S_{i}\right) _{\ell k} }_{=R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}}$
(since $N_{i}$ is the matrix $M_{q}$ without column $i$, while $S_{i}^{\prime }$ is the matrix $S_{i}$ without row $i$)
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} }\left( M_{q}\right) _{j\ell}\left( R_{\ell k}-\dfrac{v_{\ell}}{v_{i} }R_{ik}\right) $
$=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }\left( R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}\right) $
(here, we added an $\ell=i$ addend to the sum; this did not change the sum because this addend is $0$)
$=\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}R_{\ell k}}_{\substack{=\left( M_{q}R\right) _{jk}=\delta _{jk}\\\text{(since }R\text{ is a}\\\text{right inverse of }M_{q}\text{)} }}-\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}\dfrac{v_{\ell}}{v_{i}}R_{ik}}_{=\dfrac{1}{v_{i}}R_{ik}\sum_{\ell \in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}}$
$=\delta_{jk}-\dfrac{1}{v_{i}}R_{ik}\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}} _{\substack{=0\\\text{(by \textbf{(2)})}}}=\delta_{jk}$.
This shows that $N_{i}S_{i}^{\prime}=I_{n-1}$, and thus $S_{i}^{\prime} =N_{i}^{-1}$ (since $N_{i}$ and $S_{i}^{\prime}$ are $\left( n-1\right) \times\left( n-1\right) $-matrices).
Now, we recall Cramer's rule for the inverse of a matrix. It essentially says that the inverse of a square matrix is obtained by dividing its adjoint by its determinant. In other words, if $X$ is an invertible $p\times p$-matrix for some $p\in\mathbb{N}$, and $j$ and $k$ are elements of $\left\{ 1,2,...,p\right\} $, then
(3) $\left( X^{-1}\right) _{jk}=\dfrac{1}{\det X}\left( -1\right) ^{j+k}\det\left( X\text{ without row }k\text{ and column }j\right) $.
Now, recall that $S_{i}^{\prime}=N_{i}^{-1}$. Hence,
$\left( S_{i}^{\prime}\right) _{jk}=\left( N_{i}^{-1}\right) _{jk} =\dfrac{1}{\det\left( N_{i}\right) }\left( -1\right) ^{j+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ (by (3))
$=\dfrac{1}{\left( -1\right) ^{q+i}v_{i}}\left( -1\right) ^{j+k} \det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ (since $\det\left( N_{i}\right) =\left( -1\right) ^{q+i}v_{i}$)
$=\dfrac{1}{v_{i}}\left( -1\right) ^{i+j+q+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $
for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$. In other words,
(4) $v_{i}\left( S_{i}^{\prime}\right) _{jk}=\left( -1\right) ^{i+j+q+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $
for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$.
Let $\mathbf{B}$ be the (unique) increasing bijection from $\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $ to $\left\{ 1,2,...,n-1\right\} $. Then, for all $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, we have
(5) $v_{i}\left( S_{i}\right) _{jk}=\left( -1\right) ^{i+\mathbf{B} \left( j\right) +q+k}\det\left( M_{q}\text{ without row }k\text{ and columns }i\text{ and }j\right) $.
(Indeed, this follows from (4), applied to $\mathbf{B}\left( j\right) $ instead of $j$, because $\left( S_{i}^{\prime}\right) _{\mathbf{B}\left( j\right) ,k}=\left( S_{i}\right) _{jk}$ (since $S_{i}^{\prime}$ is the matrix $S_{i}$ without row $i$) and because column $j$ of $M_{q}$ is column $\mathbf{B}\left( j\right) $ of $N_{i}$ (since $N_{i}$ is the matrix $M_{q}$ without column $i$).)
Now, fix $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $. We need to show that $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$. We WLOG assume that $j\neq i$ (since otherwise, $C_{qi}R_{jk}-C_{qj}R_{ik} =0$). Thus, $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $. Since $C_{qi}=v_{i}$ and $C_{qj}=v_{j}$ (by the definitions of $v_{i}$ and $v_{j}$), we have
$\underbrace{C_{qi}}_{=v_{i}}R_{jk}-\underbrace{C_{qj}}_{=v_{j}}R_{ik} =v_{i}R_{jk}-v_{j}R_{ik}=v_{i}\underbrace{\left( R_{jk}-\dfrac{v_{j}}{v_{i} }R_{ik}\right) }_{\substack{=\left( S_{i}\right) _{jk}\\\text{(by the definition of }S_{i}\text{)}}}$
$=v_{i}\left( S_{i}\right) _{jk}=\left( -1\right) ^{i+\mathbf{B}\left( j\right) +q+k}\det\left( M_{q}\text{ without row }k\text{ and columns }i\text{ and }j\right) $
(by (5)),
which is obviously a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$ (and independent of the choice of $R$), qed.
Thanks for a very nice question, and sorry for this mess of an answer...