If $3 \not \in S$ then the answer to Question($S$) is yes.
There are integral commutative fusion rings which are not of Frobenius type.
Examples:
- Non-simple: rank $4$, global dimension $15$, type $[1,1,2,3]$, fusion rules:
$$ \begin{smallmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{smallmatrix} , \ \begin{smallmatrix} 0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1 \end{smallmatrix} , \ \begin{smallmatrix} 0&0&1&0\\0&0&1&0\\1&1&1&0\\0&0&0&2 \end{smallmatrix} , \ \begin{smallmatrix} 0&0&0&1\\0&0&0&1\\0&0&0&2\\1&1&2&1 \end{smallmatrix} $$
- Simple: rank $6$, global dimension $143$, type $[1,4,4,5,6,7]$, fusion rules:
$$ \begin{smallmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0\\1&0&1&1&1&0\\0&1&0&1&0&1\\0&1&1&1&0&1\\0&1&0&0&1&2\\0&0&1&1&2&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0\\0&1&0&1&0&1\\1&0&2&0&0&1\\0&1&0&2&1&0\\0&0&0&1&2&1\\0&1&1&0&1&2\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0\\0&1&1&1&0&1\\0&1&0&2&1&0\\1&1&2&1&0&1\\0&0&1&0&2&2\\0&1&0&1&2&2\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0\\0&1&0&0&1&2\\0&0&0&1&2&1\\0&0&1&0&2&2\\1&1&2&2&1&1\\0&2&1&2&1&2\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1\\0&0&1&1&2&1\\0&1&1&0&1&2\\0&1&0&1&2&2\\0&2&1&2&1&2\\1&1&2&2&2&2\end{smallmatrix} $$
Note that $15= 3 \times 5$ and $143 = 11 \times 13$. They admit no (pseudo)unitary categorification because by MR2098028, any fusion category of Frobenius-Perron dimension $pq$ (with $p,q$ different odd primes) is group-theoretical, whereas by MR2735754, a (weakly) group theoretical fusion category is of Frobenius type.
Note that pseudounitary means that Frobenius-Perron and global dimensions coincide. Moreover unitary implies pseudounitary.