If I've understood your question correctly, you're right that $C(q,f)\not=0$ always and there is a natural representation-theoretic proof of this result (before I start let me say that I don't know how to get these gothic $q$s and $N$s as in your question, so I am just using usual $q$s and $N$s, but they are ideals of $F$ just like yours). The one thing I am worried about is that I do not know what a "numerical character" is, probably because I think about Hilbert modular forms in a different way to you. For me, a Hilbert modular form is really just an automorphic representation of $GL(2,F)$ with certain properties, and the natural generalisation of the character of a classical modular form in this setting is the the following construction. Take the central character of this representation, which is a character of the ideles of $F$. This character decomposes as the product of a power of the norm character and a finite order character, and this finite order character is the natural generalisation of the character of the form. For me, the theorem is that if the conductor of $f$ is $N$, if $q^t$ is the exact power of $q$ dividing $N$, and if $q^t$ is also the exact power of $q$ dividing the conductor of the finite order character, then $C(q,f)\not=0$, where in this generality I am interpreting that as saying that the local $L$-factor attached to the automorphic representation at $q$ is $(1-c.Norm(q)^{-s})^{-1}$ with $c\not=0$.

So, as the referee states, this statement can be proved purely representation-theoretically. It is also a purely local assertion in fact. The automorphic representation is a tensor product of local representations and the local $L$-factor at $q$ can be computed from knowing $\pi_q$, the factor at $q$. So we are done by the following purely local theorem, where now $K$ is the completion of the totally real field $F$ at the prime $q$:

Thm) Say $K$ is a finite extension of $\mathbb{Q}_p$, say $\pi$ is a smooth admissible irreducible representation of $GL(2,K)$, say $\pi$ is ramified, has conductor $\pi^t$ ($\pi$ a uniformiser of $K$, so $\pi$ is my old $q$), and say the central character of $\pi$ also has conductor $\pi^t$. Then $\pi$ is a ramified principal series representation associated to two character, one unramified and one ramified of conductor $t$.

The reason the result you want follows is that the $L$-function of $\pi$ is $(1-c.Norm(\pi)^{-s})$ with $c$ equal to the value at a uniformiser of the unramified character.
These sorts of assertions (explicit computations of $L$-functions) can all be found in Jacquet-Langlands, a book which changed my life, but I am sure that there are references which are a gazillion times more readable nowadays.

So now all we have to do is to prove the theorem. Well there are probably purely representation-theoretic arguments, but I don't know them, so I am going to use the following trick: hit everything with local Langlands. This translates the result we want into a question about 2-dimensional representations rather than infinite-dimensional ones, so we'll be in much better shape. To make this part of the argument work you need to have an explicit hold on what local Langlands says for $GL(2)$.

OK so apply local Langlands to $\pi$ and we get a Weil-Deligne representation $(\rho,N)$ of the Weil group of $K$. And we know that the conductor of this representation is $\pi^t$ and the conductor of its determinant is also $\pi^t$, and we want to prove that $\rho$ is reducible with one ramified and one unramified character on the diagonal, and that $N=0$. Then we're done.

OK so first I'll show $N=0$. This is because if $N\not=0$ then the definition of a Weil-Deligne representation forces $\rho$ to be $\chi+\chi|.|$ with $|.|$ the norm character. And we now compute conductors. If $\chi$ is unramified then the conductor of $(\rho,N)$ is $\pi$ but the determinant is unramified, so our hypotheses do not apply (this the situation for elliptic curves with multiplicative reduction, for example; curve has bad reduction but character is unramified at $q$). And if $\chi$ is ramified and has conductor $\pi^s$ with $s\geq1$ then $\rho$ has conductor $\pi^{2s}$ so again we can't be here because $s\not=2s$.

It remains to deal with the $N=0$ case. We have a representation $\rho$ with some conductor $\pi^t$ and its character also has conductor $\pi^t$. Say first that $\rho$ is the sum of two characters $\sigma_1$ and $\sigma_2$ of conductors $t_1$ and $t_2$. Then the conductor of $\rho$ is $t_1+t_2$ and the conductor of its determinant is at most the max of $t_1$ and $t_2$, so if these are equal then one of the $t_i$ had better be zero, and so the other one had better be non-zero, and this is the case that is really happening.

All that is left now is the case where $\rho$ is irreducible. [Edit: removed incomplete answer and replaced it with complete one]. To do this case one just looks at the definition of the conductor of a representation. It's a sum of the form $\sum_i c_i.\dim(V/V^{G_i})$
where the $c_i$ are rational and the $G_i$ are running through a filtration on the inertia subgroup. The moment some $V^{G_i}$ is zero then you're in trouble, because then the sum contributes 2 to the conductor of $\rho$ and at most 1 to the conductor of its determinant. So each $V^{G_i}$ had better have dimension 1 or 2. In particular there are some inertial invariants. But these form a Galois-stable subspace, so the irreducible case cannot happen and we are finally done!