Looking at examples which involve classical groups, I'd say that the answer to your question is no.  A useful source to consult is the article on "Conjugacy classes" by Springer and Steinberg in the 1970 Lecture Notes in Mathematics No. 131 (IAS seminar write-ups), especially part IV.2 of the article.    Here they describe for classical groups explicit Levi factors in centralizers of unipotent or nilpotent elements (which are mostly interchangeable in characteristic 0).    For example, if you start with a symplectic group (simply connected), you typically get a Levi factor which is a product of various symplectic and special orthogonal groups.    The latter are not simply connected, however.   
(In most of this discussion, the characteristic of the field isn't important.)

P.S. It's usually a delicate matter in a given *simply connected* semisimple group $G$ to decide when the derived subgroup of some Levi subgroup in a proper connected subgroup of $G$ is itself simply connected.   (Even leaving aside in prime characteristic the more delicate question of whether such Levi subgroups exist.)   On the positive side, parabolic subgroups are well-behaved: they always have Levi decompositions (with all Levi subgroups conjugate in $G$) and the derived subgroup of such a Levi is always simply connected.    But the study of various centralizers in $G$ is usually not at all straightforward.  

I'm emphasizing arbitrary characteristic to underline the fact that for algebraic groups the concept *simply connected* is usually understood in terms of the position of a root lattice in a weight lattice rather than just in terms of covering groups.   For example, in their paper *Groupes reductifs*
and the later complements, Borel-Tits constructed a split group in a certain way inside a nonsplit group; but even when the original group is simply connected it requires a non-obvious argument with roots and weights to see that their split subgroup is simply connected.