Finitely generated matrix groups whose eigenvalues are all algebraic

Here's one easy example. Let $G$ be generated by $\pmatrix{1 & x\cr 0 & 1}$ for $x$ in some finite set $X$ of complex numbers. All eigenvalues are $1$, so we can take $k = \mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of $GL(2,\mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).

At the positive side, if $G$ acts irreducibly on $\mathbf{C}^n$ and $k$ is an arbitrary subfield of $\mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(\mathbf{C})$ (keeping the irreducibility assumption).

See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)

Robert Israel's simple example shows that some assumption such as irreducibility has to be done.

Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${\rm GL}(n,\mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.

Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.

In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${\rm GL}(n,\mathbb{C})$ is conjugate within ${\rm GL}(n,\mathbb{C})$ to a subgroup of ${\rm GL}(n,\mathbb{Q}[\omega]),$ where $\omega$ is a primitive complex $|G|$-th root of unity.