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. Then so is $A=k+I$?

More explicitly, choose $v_1$, $\dots$, $v_m$ in $R$ so that their images 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 $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$.
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 $1$, $v_1$, $\dots$, $v_m$, $x_1$, $\dots$, $x_n$ plus a polynomial in the $x_i$.
And the elements of $A$ can thus be written as a polynomial in the $x_i$.