The answer is yes, even in the $2 \times 2$ case.  Let $q_1,q_2,\ldots$ be an enumeration of the rational numbers.  Let $Q_j$ be the closed interval $[q_j-1/j,q_j+1/j]$.  Let $I_0=[0,2\pi]$.  Let $z=2e^{i \theta}$ for a $\theta \in I_0$ to be determined.

By induction, we construct positive integers $n_1 < n_2 < \ldots$ and closed intervals $I_0 \supseteq I_1 \supseteq \cdots$ such that for each $j$, the trace $z^{n_j} + \bar{z}^{n_j}$ is in $Q_j$ whenever $\theta \in I_j$.  Namely, if $n_1,\ldots,n_{j-1},I_1,\ldots,I_{j-1}$ have been determined already, then for any sufficiently large $n_j$, the set of $\theta$ such that $z^{n_j} + \bar{z}^{n_j}$ is in $Q_j$ is a union of closed intervals such that every real number is within $2\pi/n_j$ of a point inside this union and within $2\pi/n_j$ of a point outside this union, so if $n_j$ is chosen large enough, one such interval in this union will be completely contained in $I_{j-1}$ and we name it $I_j$.

The intersection of a descending chain of closed intervals is nonempty, so we can choose $\theta$ such that $\theta \in I_j$ for all $j$.  Then $\lbrace z^n+\bar{z}^n : n \ge 1 \rbrace$ contains an element of $Q_j$ for each $j$, so it is dense in $\mathbb{R}$.