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I'm not sure I understand why this would be a more general statement. If $A=B^{op}$ then the hypotheses look the same but the conclusions are different. I'm also not sure what the purpose of the ${}^{op}$ is.
You need the result that every finite subgroup of the multiplicative group of a field is cyclic. The easiest way to prove this is by using that a polynomial of degree $d$ has at most $d$ roots and that a finite abelian group is cyclic if and only if it's order is its exponent.
I don't know what non-singular means either but perhaps one can start with a non-singular non-injective module. If its injective hull is also non-singular we see that the answer is yes.
If you look closely you'll see that I do construct the cuspidal representations after restriction to $B$ though not on the whole of $GL_2(\mathbb{F}_q)$. They are constructed exactly as Paul Broussous indicates in his answer but more details are provided. You can see from the characters that the $\mu_\theta$ which are constructed as representations coincide with the restrictions of the cuspidal representations.
You can find a construction in chapter 9 of my notes dpmms.cam.ac.uk/~sjw47/2023Lectures.pdf. More precisely you want the things I call $\mu_\theta$ at the end of section 9.3.
