Bjorn has already answered this question in the affirmative, and shown that such matrices do exist. I'd like to add a further comment here though - 'almost no' matrices satisfy the required property. That is, the collection of 2x2 matrices such that Tr(C^n) is dense in R has zero Lebesgue measure.
We know that Tr(C^n) = a^n + b^n where a,b are the roots of the characteristic polynomial of C. If a and b are both real then it is not possible for C to have the required property. The only possibility is where they are complex conjugates, a = r exp(iθ), b = r exp(-iθ) for r >1. Then, Tr(Cn)=2rcos(nθ). Suppose that θ is uniformly distributed over [-π,π], so that exp(inθ) is uniformly distributed on the unit circle for each n. For any positive K, |Tr(C^n)|<K is equivalent to |cos(nθ)|<r-nK/2. The set of values of exp(inθ) for which this holds forms a pair of arcs of length r -nK (to leading order). So,
$$\mathbb{P}(\vert{\rm Tr}(C^n)\vert\lt K)\approx r^{-n}K/\pi$$
to leading order. Summing over n, this is finite. Then, the Borel-Cantelli lemma says that, with probability one, |Tr(Cn)|<K only finitely often. So, with probability 1, |Tr(Cn)| diverges to infinity.