As already pointed out by Emil Jeřábek in the comments, this is Hilbert's Tenth Problem over the rationals, whose computability is a notorious open question.
In this case, note that the Boolean formula $f$ is satisfiable iff $t(f)$ has an integer zero: a satisfying assignment, when treated as 0s and 1s, gives an integer zero, and conversely, given an integer zero $\vec{\alpha}$, taking it mod 2 gives a satisfying Boolean assignment (this follows from the form of the translation $t(\bullet)$). Now, given a rational solution to $t(f)=0$, if all of the denominators involved are odd, then we may still take it mod 2 to get a Boolean solution, so we have
Boolean solution iff integer solution iff rational solution w/ odd denominators
Therefore, deciding the existence of integer roots or of rational roots with odd denominators for polynomials of the form $t(f)$ is $\mathsf{NP}$-complete.
This leaves us with the question of how hard it is to decide existence of a rational root of $t(f)$ in general (that is, without restriction on the denominators). I do not know anything more about this question than about Hilbert's Tenth Problem over $\mathbb{Q}$ (addressed in Q1).