The answer is Yes.

The generating function for $t_n$ is
$$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$
Correspondingly,
$$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$
It follows that for $n>0$,
$$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv (-1)^{\lfloor n/2\rfloor}\binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$
from where [Lucas' theorem](https://en.wikipedia.org/wiki/Lucas%27s_theorem) gives the desired result.