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 $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv \binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$ from where Lucas' theorem gives the desired result.