Your question may be recast as:
Let $A$ be an integral matrix $A\in SL(n,\mathbb{Z})$. Is it true that there exists $N(n)$ such that there is $k\leq N(n)$ with $tr(A^k)>2 $ (with a few small exceptional cases)?
Let the singular values of $A$ be denoted $\lambda_1,\ldots, \lambda_n$. Then $tr(A^k)=\sum_{i=1}^n \lambda_i^k =p_k(\lambda_1,\ldots,\lambda_n)$, where $p_k$ is the power symmetric polynomial of degree $k$ in $n$ variables. Now, we observe that $p_k$ is a polynomial in $p_1,\ldots, p_n$ for $k>n$. I found another mathoverflow question which computed the first few of these.
Consider $n=2$ (note, $SL(2)=Sp(2)$), and suppose $p_i(\lambda_1,\lambda_2) \leq 2, 1\leq i\leq 2$. It turns out that if $-2\leq p_1 \leq 2$, then $|tr(A^k)|\leq 2$ for all $k$, and $A$ is either parabolic or elliptic (and similarly if $|p_2|\leq 2$). So assume $p_1,p_2 \leq -3$. Then $p_3=-p_1^3 +\frac32 p_1p_2 \geq \frac{81}{2}$. Thus, $N(2)$ exists.
Now, consider $n=3$. One has the relation $p_4=\frac16 p_1^4−p_1^2p_2+\frac12 p_2^2+\frac43 p_1p_3$. Suppose that $p_1\leq -R$, for some $R>>0$, and $p_2,p_3\leq 2$. Then one has $p_4 = \frac16 p_1^4-p_1^2p_2 +\frac12 p_2^2+\frac43 p_1p_3 \geq \frac16 R^4 -2R^2 -\frac43 2R$. Then choose $R$ large enough that $\frac16 R^4 -2R^2 -\frac83 R >2 $, then one has $p_4 >2$ if $p_1\leq -R$. Now, suppose $p_2\leq -R$ or $p_3 \leq -R$. Then we similarly conclude that $p_8 >2$ or $p_{12} >2$. Otherwise, $ -R\leq p_i\leq 2$ for $1\leq i\leq 3$, and there are only finitely many characteristic polynomials $det(xI-A)$ of such matrices, since the elementary symmetric polynomials are determined by the power sum symmetric polynomials. Thus, there is a uniform $N(3)$ such that for some $k\leq N(3)$, $tr(A^k)>2$.
Similarly, for $n=4$, $p_5=−\frac{1}{24} p_1^5+\frac{5}{12} p_1^3p_2−\frac58 p_1p_2^2−\frac56 p_1^2p_3+\frac56 p_2p_3+\frac54 p_1p_4$. The leading order term here is $-\frac{1}{24} p_1^5$, so if $p_1\leq -R$ for $R>>0$, and $p_2,p_3,p_4\leq 2$, one concludes that $p_5 \geq \frac{1}{24} R^5 -\frac{5}{6} R^3 +\frac58 R p_2^2 -\frac56 R^2 p_3 +\frac56 p_2 p_3 - \frac52 R$. If $p_3 < 0, 0<p_2<2$, then the term $\frac56 p_2p_3$ will be dominated by the term $-\frac56 R^2 p_3$ for $R>>0$. If $p_2 <0, 0<p_3<2$, then $\frac56 p_2p_3$ will be dominated by $\frac58 R p_2^2$, and otherwise $\frac56 p_2 p_3 \geq 0$. So we conclude that for $R>>0$ and $p_1\leq -R$, $p_5>2$. Similar to before, we may replace $p_1$ with $p_2, p_3$, or $p_4$ if any of these is $\leq -R$, to conclude that $p_{5i}>2$ for $i=2,3$ or $4$, and otherwise there will be only finitely many possibilities. So $N(4)$ exists.
There's a similar pattern to the general formula, which is given in a comment by Darij Grinberg to this question. He gives the formula: $$ \sum_{i=1}^{n+1} \frac{(-1)^i}{i!} \sum_{j_1,\ldots,j_i\geq 1; j_1+\cdots+j_i=n+1} \frac{1}{j_1 \cdots j_i} p_{j_1}\cdots p_{j_i}= 0.$$ This gives the same leading order asymptotics $p_{n+1}=(-1)^np_1^{n+1} + O(p_1^n)$. A similar analysis might yield therefore a bound for all $n$, but it might take some care to work it out.