Since the image of $\varphi$ lies inside of $\mathbb{Z}[x_1,\ldots,x_n]$, you can replace $\mathbb{Q}[x_1,\ldots,x_n]$ with $\mathbb{Z}[x_1,\ldots,x_n]$ (since the image vanishes in the latter if and only if it vanishes in the former).
Any map from $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ to a commutative ring factors uniquely through $\mathbb{Z}[x_1,\ldots,x_n]$ by the universal property of the polynomial ring.
Explicitly: given any ring homomorphism $f\colon R\to S$, where $R$ and $S$ are commutative, and given any elements $s_1,\ldots,s_n\in S$, there exists a unique homomorphism $\overline{f}\colon R[x]\to S$ such that $\overline{f}(r)=f(r)$ for all $r\in R$ and $\overline{f}(x_i) = s_i$ for $i=1,\ldots,n$.
In particular, if $\Phi$ is any map from $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ to a commutative ring $S$, then $\Phi(x_i)$ determines a unique homomorphism $\psi\colon\mathbb{Z}[x_1,\ldots,x_n]\to S$ with $\psi(x_i)=\Phi(x_i)$ and $\psi(1) = \Phi(1)$. This map makes a commuting triangle with the quotient map $\varphi$. Hence we get $\Phi=\psi\circ\varphi$, as you suggest in point (1) (after replacing $\mathbb{Q}$ with $\mathbb{Z}$.
Thus, if $\Phi$ is any map from $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ to a commutative ring $S$, then necessarily $\mathrm{ker}(\varphi)\subseteq\mathrm{ker}(\Phi)$, which answers your first question.
(Or, you can think about the "multidegree" of a monomial in $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ which only counts the total exponent of each variable; an element of $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ lies in the kernel of $\varphi$ if and only if it is a sum of "homogeneous terms", i.e., same multidegree for all monomials, where the coefficients add up to $0$. Any such homogeneous term will also lie in the kernel of any $\Phi$, since the image is commutative).
But I'm not sure why you would need to go through this for your second question: $\mathrm{trace}(AABABB-AABBAB)$ can be written as a (commutative) polynomial in the $8$ variables that form the entries of $A$ and $B$, with integer coefficients; i.e., an element of $\mathbb{Z}[x_1,\ldots,x_n]$ (not of $\mathbb{Z}\langle A,B\rangle$, because we are looking at the trace, not the product itself). If $3$ times this expression is $0$ as an element of $\mathbb{Z}[x_1,\ldots,x_n]$, then this expression is itself $0$ simply because $\mathbb{Z}[x_1,\ldots,x_n]$ is an integral domain. And from there, you get that the corresponding expresssion is also identically zero over any field, because it is zero in the initial polynomial ring $\mathbb{Z}[x_1,\ldots,x_n]$.
That is, the lack of commutativity between the matrices does not matter here, because the expression we are looking for, the trace, is computed in terms of the coefficients of the matrices via commutative (though complicated) operations.
E.g., if you were talking about the trace of $AB-BBA$, you would have $\begin{align*} A&=\left(\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array}\right),\\ B&=\left(\begin{array}{cc} b_{11} & b_{12}\\ b_{21} & b_{22} \end{array}\right),\\ \mathrm{trace}(AB - BBA) &= \mathrm{trace}(AB) - \mathrm{trace}(BBA)\\ &= \Bigr(a_{11}b_{11} + a_{12}b_{21} + a_{21}b_{12}+a_{22}b_{22}\Bigr) \\ &\qquad\mathop{+} \Bigl( (b_{11}^2+b_{12}b_{21})a_{11} + (b_{11}b_{12}+b_{12}b_{22})a_{21}\\ &\qquad\quad\mathop{+} (b_{21}b_{11}+b_{22}b_{21})a_{12} + (b_{21}b_{12}+b_{22}^2)a_{22}\Bigr) \end{align*}$ This is a polynomial in $\mathbb{Z}[a_{11},\ldots,b_{22}]$, with commuting variables; not an element of $\mathbb{Z}\langle a_{11},\ldots,b_{22}\rangle$.
Note that you cannot interpret the trace map as a ring homomorphism from $\mathbb{Z}\langle A,B\rangle$ to something, because the trace is not a ring homomorphism: the trace of the product is not the product of the traces.