Yes.
Observe that $\mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(\mathrm{gr}(A/I)) = \oplus_{i=0}^r~\mathrm{gr}(A/I)_i$.
Further observe that $$\dim F_i(\mathrm{gr}(A/I) = \sum_{i=0}^r \dim \mathrm{gr}(A/I)_i = \sum_{i=0}^r (\dim F_i(A/I) - \dim F_{i-1}(A/I)) = \dim F_i(A/I).$$
Therefore $\mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.