Is there an NP-complete problem, such that the decision version of its counting problem is not PP-complete?
Putting your question in more precise terms, you ask whether the following claim holds:

$* \hspace{1mm} R(x,y) \text{ is an NP-complete relation}\Rightarrow count_R \text{ is PP-complete}$

Where $count_R$ is defined as follows:

$count_R=\left\{(x,k) \big| \hspace{1mm} \left\lvert\left\{y : R(x,y)\right\}\right\rvert\ge k\right\}$.

We call a relation $R(x,y)$ NP-complete if it is computable in polynomial time, and the language it defines $L_R=\left\{x | \exists y \hspace{1mm} R(x,y)=1\right\}$ is NP-complete.

We talk in terms of relations, since as you mentioned, the counting version has to be defined relative to some specific verifier.

It seems that this is an open question, as (*) implies: 

$** \hspace{1mm} R(x,y) \text{ is an NP-complete relation}\Rightarrow \#R\text{ is #P-complete}$

Where $\#R(x)=\left\lvert \left\{y : R(x,y)\right\} \right\rvert$.

To see why * implies the above, let $R(x,y)$ be some NP-complete relation. Using (*), $count_R$ is PP complete, so $count_{\text{SAT}}\le_p count_R$. In that case, $\#SAT\in\mathsf{FP}^{\# R}$ and thus $\#R$ is $\# P$-complete (use binary search, where in each cutoff you apply the reduction from $count_{\text{SAT}}$ to $count_R$ and query the $\#R$ oracle on the result).

To my knowledge, (**) is currently open. See this related question from cstheory. Also related.