Let $A(x)=1+xB(x) \in \mathbb{C}[[x]]$ be a power series. It is a standard fact that $A(x)$ is a rational function iff there is a finite number of complex numbers $\alpha_i,\beta_j$ such that the $n$-th coefficient of $$x \frac{A'(x)}{A(x)}$$ is of the form $\sum a_i^n - \sum b_j^n$. If we write $A(x)$ as an infinite product $$A(x) = \prod_{n \ge 1} (1-x^n)^{-a_n}$$ for some $a_n \in \mathbb{N}$ (this is always possible, in a unique way), we have $$x \frac{A'(x)}{A(x)} = \sum_{n \ge 1} (\sum_{d \mid n} a_d d)x^n.$$
In our case, $a_{2n} = SRMI_q(2n), a_{2n-1} = 0$. I will focus on the even $q$ case, for simplicity.
Let $n$ be a positive integer. We have
$$\sum_{d \mid n} a_d d =\sum_{2d \mid n} a_{2d} 2d=\sum_{m \mid n/2} q^m \sum_{i \mid n/(2m), \text{ odd}} \mu(i).$$ The identity $\sum_{i \mid s}\mu(i)=1_{s=1}$ implies that $\sum_{i \mid n/(2m), \text{ odd}} \mu(i) = 1_{n/{2m} \text{ is a power of 2}}$, and thus, if we let $2^{v_2(n)}$ be the highest power of $2$ dividing $n$, then $$(*) \sum_{d \mid n} a_d d=\sum_{i=0}^{v_2(n)-1} (q^{2^{-(i+1)}})^n.$$
It is not hard to see that the sum $(*)$ cannot be of the form $\sum a_i^n - \sum b_j^n$. (Sketch of Proof: The values of $\alpha_i,\beta_j$ are the poles of $x\frac{A'(x)}{A(x)}$, while $(*)$ implies that $x\frac{A'(x)}{A(x)}$ has infinitely many poles at $q^{2^{-(i+1)}}$.)
Identity $(*)$ also implies that $$A(x) = \prod_{i \ge 1}(1-qx^{2^i})^{-2^{-i}}$$ which leads to $$A(x^2) = \Big( A(x)(1-qx^2)^{1/2})^2.$$ As for the odd $q$ case, the same arguments lead to $$A(x) = \prod_{i \ge 1}(\frac{1-qx^{2^i}}{1-x^{2^i}})^{-2^{-i}}, A(x^2) = \Big( A(x)(\frac{1-x^2}{1-qx^2})^{1/2})^2.$$