Let $\mathbb{F}$ be a field and consider the free algebra $\mathbb{F}\langle x_1,\ldots, x_n \rangle $, that is, the algebra of non-commutative polynomials with coefficients from $\mathbb{F}$. Let $p\in\mathbb{F}\langle x_1,\ldots, x_n \rangle $.
Question: Is it possible that there exist two $\mathbb{F}$-algebras $A_0$ and $A_1$ and an assignment $b\in\{0,1\}^n$ of $0,1$ values (the zero and unit elements in $A_0$ and in $A_1$, respectively) to the $x_i$ variables such that $p(b)=0$ when computed in $A_0$ but $p(b)\neq 0$ when computed in $A_1$?
For any $\mathbb{F}$-algebra $A$ the evaluation map $\mathbb{F}\langle x_1,\dots,x_n \rangle \rightarrow A$ at $b$ factors through $\mathbb{F} \rightarrow A$ because $b \in \mathbb{F}^{n}$ by your assumption, that is, it is the composition of the evaluation map $\mathbb{F}\langle x_1,\dots,x_n \rangle \rightarrow \mathbb{F}$ with the canonoical map $\mathbb{F} \rightarrow A$. Hence $p(b)$ is computed in $A$ by computing it in $\mathbb{F}$ and taking the image under $\mathbb{F} \rightarrow A$ which is an injection. So $p(b) = 0$ in $A$ if and only if this is the case in $\mathbb{F}$ and so this identity is independent of the choice of $A$.