My immediate instinct was "no", and I tried the system of short roots in $\mathsf C_2$. This is almost a counterexample, but not quite: there's no semisimple subalgebra of $\mathfrak{sp}_4$ normalised by "the" Cartan subalgebra whose roots are the short roots, but $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$ is a subalgebra whose root system has the same type as the system of short roots. (This was mentioned by @GrantB. in a comment.) Similarly for $\mathfrak g_2$: there's no semisimple subalgebra normalised by "the" Cartan subalgebra whose roots are the short roots, but the subalgebra generated by the long root spaces is $\mathfrak{sl}_3$, which is of the same type.
I am visiting Jeff Adler and mentioned your question to him, and he pointed out that I was just being too conservative with rank: if instead of $\mathsf C_2$ one looks at $\mathsf C_3$, then the subsystem of short roots is of type $\mathsf D_3$, but there is no $\mathfrak{so}_6$ subalgebra of $\mathfrak{sp}_6$. (For example, one can verify this using Borel–de Siebenthal theory.) Since this is Jeff's answer, I have made it community wiki to avoid reputation. If you like it, then please go up-vote one of his questions or answers.
If, however, you require additionally that the sub-root system be closed, in the sense that any integral linear combination of roots of $\mathfrak n$ that is a root of $\mathfrak g$ is also a root of $\mathfrak n$, then, as mentioned by @მამუკაჯიბლაძე and @GrantB., the subalgebra of $\mathfrak g$ generated by the appropriate root spaces with respect to a fixed Cartan subalgebra of $\mathfrak g$ is a semisimple subalgebra of the desired type. (I mention, by the way, that it's very difficult for me to think of an algebra denoted by $\mathfrak n$ as semisimple; my brain can't but see it as the nilradical of a parabolic subalgebra!)