Concerning 

> When not, are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is 
*split* provided $\operatorname{End}_A(S) = k$
for every simple $A$-module $S$. 

[This terminology is consistent with that used
in other contexts -- $A$ is split
just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$ (indeed, since $k \subset \ell$ is separable and since $A \otimes_k \ell$ is split,  $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra whose center is thus a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$). 

Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. 
Then $A \otimes_k \ell$ is split
for an *Abelian* extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).