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. It only shows that the hypotheses are reasonable.



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 $R$ is generated as an $A$-module by $1$, $v_1,\dots,v_m$. So $A$ is a finitely generated $k$-algebra by the 
Artin-Tate lemma in <a href="http://en.wikipedia.org/wiki/Artin–Tate_lemma">wikipedia</a>. 


One may argue more directly. Every $r\in R$ may be written as a $k$ linear combination of $1$, $v_1,\dots,v_m$ plus an element, say $f(r)$, in $I$.
Now take a generating set $y_1,\dots,y_m$ of the $k$-algebra $R$ and choose $x_1$, $\dots$, $x_n$ in $I$ so that the $f(y_i)$ and the $f(v_iv_j)$
are amongst the $x_j$.

We claim that 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 $x_j$, $v_ix_j$.
Indeed it is easy to check that the set of elements that can be written this way is invariant under multiplication by the $v_i$ and the $x_j$, hence also
by the $y_j$.

If an element of $A$ is written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$  plus a polynomial in the $x_j$, $v_ix_j$,
then it must be a polynomial in the $x_j$, $v_ix_j$.

So the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$.