Let $\Gamma : \mathcal A \to \mathcal B$ be an equivalence of categories where $\mathcal A$ is an exact category, $\mathcal B$ is an additive full subcategory of the category of $R$-modules for some commutative Noetherian ring $R$ , and $\Gamma$ is an additive functor .
Then is it true that $\Gamma$ induces an isomorphism of Grothendieck groups $K_0(\mathcal A)\cong K_0(\mathcal B)$ ? If this is not true in general, would some extra assumptions on the categories of the functor would make it true ?
No, not in general. $\mathcal{A}$ could be the same as $\mathcal{B}$ as an additive category (and $\Gamma$ the identity functor), but with a different exact structure. Then there is no reason to expect the Grothendieck groups to be isomorphic.
For example, let $\mathcal{A}$ be the category of finitely generated abelian groups with the exact structure given by split short exact sequences, and $\mathcal{B}$ the same category with the exact structure given by all short exact sequences. Then $K_0(\mathcal{B})\cong\mathbb{Z}$, but $K_0(\mathcal{A})$ is free abelian of countable rank.