Can every finite dimensional division ring be endormorphism ring of some representation？

Good question! The answer is no.

Let's say the characteristic of $k$ doesn't divide $G$, so that $k[G]$ is separable. Recall that a splitting field of a separable algebra $A$ over $k$ is a field extension $L$ of $k$ such that $A \otimes_k L$ is isomorphic to a product of matrix algebras over $L$. A splitting field of a finite group $G$ over $k$ is a splitting field of $k[G]$.

It's known that finite groups have abelian splitting fields; more precisely, $k[G]$ can always be split by adjoining all $|G|$-th roots of unity to $k$. (This is due to Brauer, at least in characteristic $0$.) But not all division algebras over $k$ have this property; in particular, nonabelian Galois extensions of $k$ don't. So those can't occur in $k[G]$, and hence can't occur as endomorphism rings of $G$-representations over $k$.

Edit: Separability is another constraint; if $k$ has positive characteristic, then an inseparable extension of $k$ can't occur either.

Another answer, sillier than Qiaochu's. There are fields over which there are uncountably many finite dimensional division rings. But (up to isomorphism) there are only countably many finite groups, each of which has only finitely many irreducible representations.

To make Jeremy Rickard's slick answer concrete with a specific example, let $k_0$ be any field and let $k$ be obtained from $k_0$ by adjoining uncountably many transcendental elements $x_i$. Then $k$ has uncountably many different finite field extensions: for instance, for each of the adjoined transcendental elements, you can adjoin a square root. At most countably many different division rings can come from representations of finite groups, so all but countably many of these field extensions must not.

In fact, more strongly, we can easily deduce that any finite extension which "involves" any of the $x_i$ cannot come from a representation of a finite group. More precisely, suppose $L$ is a finite extension of $k$ containing an element whose minimal polynomial does not have coefficients in $k_0$. Then the minimal polynomials of elements of $L$ involve only finitely many of the $x_i$, and so there are uncountably many permutations of the $x_i$ that give different finite field extensions of $k$. But by symmetry (since applying a permutation of the $x_i$ gives an isomorphism of the whole setup over $k_0$), if one of these extensions comes from a representation of a finite group, all of them do. Since there are uncountably many of them, this means none of them do.