Alpoge's comments oversimplify the situation slightly.
There exists a unique constant $\alpha$ modulo $p^{\lfloor t/2\rfloor}$ such that $\chi(1+x)=q^{ 2\pi i \alpha x/q}$ for $\alpha$ a mulitple of $p^{\lceil t/2 \rceil}$.
If we let $e_q(x) = e^{2\pi i x /q}$ and $F(h) = \chi( (h-a_1)(h-a_2)(h-a_3)) \ \bar{\chi}( (h-b_1)(h-b_2)(h-b_3)) e_q( C h )$ then $$F(h+x)= F(h) e_q\left( x \left( \alpha \left( \frac{1}{h-a_1} + \frac{1}{h-a_2} + \frac{1}{h-a_3} - \frac{1}{h-b_1} - \frac{1}{h-b_2}- \frac{1}{h-b_3} \right) + C \right) \right).$$
Thus the sum over the residue class vanishes unless$$ \alpha \left( \frac{1}{h-a_1} + \frac{1}{h-a_2} + \frac{1}{h-a_3} - \frac{1}{h-b_1} - \frac{1}{h-b_2}- \frac{1}{h-b_3} \right) + C =0$$ modulo $p^{\lfloor t/2 \rfloor}$.
If this rational function has only simple zeroes then it will have $O(1)$ zeroes in the $t$ even case or $O(p)$ zeroes in the $t$ odd case but with Gauss sum cancellation.
However, in other cases, where the vanishing order is higher, you can get less than square root cancellation for $t$ large. This happens when your sum locally looks like $e_q(x^3)$, for instance.
