I am just posting my comments as an answer. Without a hypothesis that the geometric generic fiber of $\pi:S\to \mathbb{A}^1_t$ is irreducible, the result is false. For a smooth compactification of $S$ on which $\pi$ extends to a morphism to $\mathbb{P}^1_t$, the finite part of the Stein factorization of $\pi$ is either an isomorphism to $\mathbb{P}^1_t$ (precisely when the geometric generic fiber is irreducible) or it is a degree-$2$ cover, i.e., a hyperelliptic curve. For appropriate choices of the coefficient polynomials $p_i(t)$, this can be any hyperelliptic curve. If the genus is $\geq 2$, then this curve has only finitely many $\mathbb{Q}$-points (by Mordell's Conjecture / Falting's Theorem), so the image in $\mathbb{P}^1_t$ is also a finite set.

However, if the geometric generic fiber is irreducible, then the compactification over $S$ is a conic bundle over $\mathbb{P}^1_t$. After base change from $\mathbb{Q}$ to some number field, this surface is rational, i.e., the surface is geometrically rational. There is a conjecture (perhaps due to Colliot-Thélène) that the set of rational points on a geometrically rational variety over a number field is dense in the Brauer subset of the set of adelic points. Assuming this conjecture, once there is a single rational point (so that the Brauer subset is nonempty), the set of rational points is Zariski dense. Thus, the image in $\mathbb{P}^1_t$ is also Zariski dense.

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