Yes. Both $R$ and $A=k+I$ are filtered by powers of $I$ and we may pass to 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 $k+I$.