The answer is no. There are uniquely determined linear polynomials $a_m(x):=x-c_m$ such that $g(x)$ is identically zero on $\mathbb{Q}$. Indeed, the condition says that, for any $n\geq 1$, $$ \sum_{0\leq m\leq n-1}(b_n-c_m)\cdot(b_n-b_1)\dots(b_n-b_m) = 0. $$ Suppose that $c_m$ has been chosen already for every $0\leq m<n-1$. Then the above equation determines $c_{n-1}$ uniquely, and we are done.

The answer is no. There are uniquely determined linear polynomials $a_m(x):=x-c_m$ over $\mathbb{Q}$ such that $g(x)$ is identically zero on $\mathbb{Q}$. Indeed, for such polynomials the condition means that $$ \sum_{0\leq m\leq n-1}(b_n-c_m)\cdot(b_n-b_1)\dots(b_n-b_m) = 0,\qquad n\geq 1. $$ Here, the second product is meant to be $1$ when $m=0$. Now suppose that, for a given $n\geq 1$, the coefficients $c_m\in\mathbb{Q}$ have been determined for $0\leq m<n-1$. Then, the above equation has a unique solution for $c_{n-1}\in\mathbb{Q}$. So we obtained a recursion for the rational numbers $c_0,c_1,c_2,\dots$, and we are done.