Yes. Both $R$ and $A=k+I$ are filtered by powers of $I$ and we may look at the associated graded rings. Now $I/I^2$ is a finitely generated $R/I$ module and $R/I$ is a finitely generated vector space, by Zariski's lemma on finitely generated $k$ algebras that are fields. So $I/I^2$ is a finitely generated vector space and the associated graded of $A=k+I$ is a finitely generated $k$ algebra. But that is not enough.
More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then choose $x_1$, $\dots$, $x_n$ in $I$ so that together with the $v_i$ they generate the $k$ algebra $R$ and so that every product $v_iv_j$ can be rewritten as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$. (First achieve the generation, then throw in more elements to achieve the rewriting. Use that for every $r\in R$ there is a $\tilde r\in I$ so that $r$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ and $\tilde r$.) Now the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$. Indeed every element of $R$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. And the elements of $A$ can thus be written as a polynomial in the $x_j$, $v_ix_j$.