If $3 \not \in S$ then the answer to Question($S$) is yes.
There is an integral commutative fusion ring which is not of Frobenius type. So we are reduced to $3 \in S$, except if the fusion ring below is one among few anomalies (which then require to be classified).
Here is an example: rank $4$, global dimension $15$, type $[1,1,2,3]$, fusion rules:
$$\left(\begin{smallmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{smallmatrix} \right) , \ \left(\begin{smallmatrix} 0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1 \end{smallmatrix} \right) , \ \left(\begin{smallmatrix} 0&0&1&0\\0&0&1&0\\1&1&1&0\\0&0&0&2 \end{smallmatrix} \right) , \ \left(\begin{smallmatrix} 0&0&0&1\\0&0&0&1\\0&0&0&2\\1&1&2&1 \end{smallmatrix} \right) $$
and character table:
$$\left[\begin{matrix} 1&1&1&1\\1&1&-1&1\\2&-1&0&2\\-2&0&0&3 \end{matrix} \right]$$
It is not of Frobenius type because $2$ does not divide $15$. Moreover it admits 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.