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 are 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 this sum cannot be of the form $\sum a_i^n - \sum b_j^n$, essentially because it depends on the $2$-adic valuation of $n$.