Here are answers by Pavel Etingof (reproduced with his authorization):
Question 1: Is there a fusion ring which is not the Grothendieck ring of a finite tensor category?
Answer: Yes, for example the rings in Example 8.19 in
https://arxiv.org/pdf/math/0203060.pdf
given by $gX=Xg=X, X^2=X+\sum_{g\in G}g$ do not admit a categorification (even non-semisimple) when $|G|>2$.
Indeed, assume the contrary. Since $gX=Xg=X$, we get that $X^2$ both projects to $g$ and contains $g$ for any $g\in G$. Thus each g and hence $\oplus_{g\in G}g$ is a direct summand of $X^2$. So any category which categorifies this ring must be semisimple. But in Example 8.19 it is shown that this ring has no semisimple categorification.
Question 2: Is there a fusion ring which is the Grothendieck ring of a finite tensor category, but not of a fusion category?
Answer: I don’t know. In positive characteristic i think it is likely that such examples exist.
Bonus question: Is there a finite tensor category over $\mathbb{C}$ whose Grothendieck ring is not a fusion ring?
Answer: absolutely. For example see the book "Tensor categories",
http://www-math.mit.edu/~etingof/egnobookfinal.pdf
p.105 (small quantum $sl(2)$). In fact, this is the typical behavior. Moreover, if the category is pivotal factorizable and non-semisimple (as in the above example), this must happen by
https://arxiv.org/pdf/1703.00150.pdf
Namely, it has a simple projective object $X$, so $X\otimes X^*$ is projective and contains projective cover $P(1)$ as a summand. If the Grothendieck ring is a fusion ring then $[X\otimes X^*:1]=1$, so since the category is unimodular (as it is factorizable), we get $P(1)=1$, hence it is semisimple.
Consider the symmetric group group $G = S_3$ of order $6$. Then $\mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1)) \cong \mathbb Z/6\mathbb Z$. Choose a generator $\omega \in \mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1))$. Then $\omega$ restricts nontrivially to every nontrivial subgroup of $G$.
It follows that $\mathbf{Vec}^\omega[G]$ is not Morita-equivalent to any other fusion category. (Recall that, for any $G,\omega$, fusion categories equivalent to $\mathbf{Vec}^\omega[G]$ are indexed by pairs consisting of a subgroup $H \subset G$ together with a 2-cochain $\psi$ on $H$ solving $\mathrm{d}\psi = \omega|_H$.)
But $G$ is noncommutative.
Best Answer
I think an answer to your question is given in Naidu, Nikshych, and Witherspoon - Fusion subcategories of representation categories of twisted quantum doubles of finite groups, theorem 1.1.
Subcategories of the double $D(G)$ are given by pairs of normal subgroups $K$, $N$ in $G$ which centralize each other, together with the datum of a bicharacter $K\times N \to \mathbb C^\times$.
So in particular if $G$ has no normal subgroups and $H$ does, then you're going to find that $D(G)$ has no nontrivial subcategories, while $D(H)$ will (one can take $K$ the normal subgroup in $H$, $N=\{id\}$, and the bicharacter $K\to \mathbb C^\times$ to be trivial, I guess).