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We fix some $t\in \Bbb C$. Let $a,b,c$ be the complex roots of the equation $$ X^3-6tX^2+18t^2 X -36 t^3\ . $$ Let $e_n$ be the elementary symmetric polynomial of degree $n$ in $a,b,c$.

Let $p_n$ be the Newton symmetric polynomial of degree $n$ in $a,b,c$.

By Vieta, $$\left\{ \begin{aligned} e_1 &= a+b+c = 6t\ ,\\ e_2 &= ab+bc+ca = 18t^2\ ,\\ e_3 &= abc = 36t^3\ . \end{aligned} \right. $$ Let $A$ be the matrix with diagonal entries $a,b,c$. Then, using the Newton identities, $$ \begin{aligned} \operatorname{Trace}(A^2) &=a^2+b^2+c^2\\ &=p_2\\ &=e_1p_1-2e_2\\ &=6t\cdot 6t-2\cdot 18t^2=0\ , \\ \operatorname{Trace}(A^3) &=a^3+b^3+c^3\\ &=p_3\\ &=e_1p_2-e_2p_1+3e_3\\ &=0-18t^2\cdot 6t+3\cdot 36t^3=0\ , \\ \operatorname{Trace}(A^4) &=a^4+b^4+c^4\\ &=p_4\\ &=e_1p_3-e_2p_2+e_3p_1-4e_4\\ &=0-0+36t^3\cdot 6t-0\\ &=216t^4\ . \end{aligned} $$ It is clear that we can chose $t$, such that the last expression is not rational. This gives and explicit counterexample.

Later edit: The general family of cubics with roots $a,b,c$, so that $p_1=p_1(a,b,c)=e_1=e_1(a,b,c)=s$ and $p_2=p_3=p_2(a,b,c) =p_3(a,b,c)=q$, thus parametrized by $s,q$, is obtained by solving $$ \begin{aligned} e_1&=p_1=s\ ,\\ 2e_2 &=e_1p_1-p_2=s^2-q\ ,\\ 3e_3 &=e_2p_1-e_1p_2+p_3=\frac 12(s^2-q)s-sq+q\ ,\\ \end{aligned} $$ and $p_4$ is obtained algebraically as $p_4=e_1p_3-e_2p_2+e_3p_1-4e_4$. (Here, $e_4=0$.) The question "It is still possible to find a counterexample so that $s$ (trace of $A$) is a natural (or even rational) number?" has a negative answer. If both $s,q$ are rational, than $p_4$ is rational, as an algebraic expression in $s,q$.