No.
Suppose we had equidistributed $(a_n)_n$ such that $\sum_{n \le t} \sin(a_n) \sim \frac{t}{\log t}$.
Then, for any $\beta \in (0,1]$, summation by parts gives $$\sum_{n \le N} \frac{\sin(a_n)}{n^\beta} = \frac{\sum_{n \le N} \sin(a_n)}{N^\beta}+\beta \int_1^N \frac{\sum_{n \le t} \sin(a_n)}{t^{1+\beta}}dt,$$ which diverges to $+\infty$ (the integral goes to $+\infty$; if $\beta < 1$, the first term does as well, and if $\beta = 1$, the first term goes to $0$).
We now construct such $(a_n)_n$. Fix $\theta = \sqrt{2}$ (any irrational would do). Let $b_n = n\theta$ mod $1$. Then $(b_n)_n$ is very equidistributed in that, for some absolute $C \ge 1$, $$\left|\sum_{n \le N} \sin(b_n)\right| \le C$$ for each $N \ge 1$ (recall $|\sum_{n \le N} e(n\theta)| \le \min(N,\frac{1}{||\theta||})$). Define $(a_n)_n = (b_1,\frac{\pi}{2},\frac{\pi}{2},b_2,\frac{\pi}{2},b_3,\frac{\pi}{2},b_4,b_5,b_6,\dots)$, where we have inserted $\frac{\pi}{2}$'s into the sequence $(b_n)_n$ at prime indices. Then we have that $(a_n)_n$ is equidistributed, since we just modified an equidistributed sequence on a density $0$ set. And due to the last centered inequality (and the density of the prime numbers), we have $\sum_{n \le t} \sin(a_n) \sim \frac{t}{\log t}$.
The main reason the answer is "no" is that equidistribution is just a sublinear assumption on $\sum_{n \le t} \sin(a_n)$. If you wanted to define a sequence $(x_n)_n$ to be $\beta$-equidistributed if $\sum_{n \le N} \sin(qx_n) = o(N^\beta)$ (rather than $o(N)$) for each $q \in \mathbb{Z}$, then the answer would be "yes", as the summation by parts formula above shows.
One last thing I want to say, at the expense of being annoying. Weyl's criterion is really the converse of what you said. It is completely trivial to show that if a sequence $(x_n)_n$ is equidistributed, then $\sum_{n \le N} \sin(qx_n) = o(N)$ for each $q$. However, the quite remarkable thing is that if $\sum_{n \le N} \sin(qx_n) = o(N)$ and $\sum_{n \le N} \cos(qx_n) = o(N)$, then $(x_n)_n$ is equidstributed. This converse is really at the heart of fourier analysis, and is still quite magical to me.
Best Answer
It's because each of the cosets of the period is equidistributed. For instance, if $p(n) = \frac{1}{2}n^2+\pi n$, then both $(p(2n))_{n \ge 1}$ and $p((2n+1))_{n \ge 1}$ are equidistributed.
Exercise: If $(x_n)_n = (p_n+y_n)_n$ where $p_n$ is periodic with period $p$ and $(y_{pk+j})_{k \ge 1}$ is uniformly distributed for each $0 \le j \le p-1$, then $(x_n)_n$ is uniformly distributed.