A simple upper bound - $2$.

Question

Let $q$ be a large prime and fix $r\in(0,\frac12)$.

Is there a $\theta\in(\frac12,1)$ such that $\Bigg|\frac{e^{2\pi iaq^r}-1}{e^{2\pi i\frac aq}-1}\Bigg|

Answer 1

No, at least not for every $a$. (The counterexample below uses $a=1$, but should, I think, generalize to $a=o(q)$.)

Namely, for the following: $r=\frac{1}{4}$, and $q=n^4+2n^3 + o(n^3)$ for a large integer $n$ (think of $n\to\infty$). Then, for $a=1$ $$ aq^r = \left(n^4+2n^3 + o(n^3)\right)^{1/4} = n+\frac{1}{2}+o\left(\frac{1}{n}\right) $$ so that the numerator is $$ \left\lvert e^{2\pi i a q^r} -1\right\rvert =\left\lvert e^{2\pi i n+i\pi+o\left(\frac{1}{n}\right)} -1\right\rvert =\left\lvert e^{i\pi+o\left(\frac{1}{n}\right)} -1\right\rvert =\left\lvert -2 + o(1)\right\rvert = 2+o(1) \tag{1} $$ while the numerator is $$ \left\lvert e^{2i\pi\frac{a}{q}}-1\right\rvert = \left\lvert 2i\pi\frac{1}{q} + o\left(\frac{1}{q}\right)\right\rvert = \frac{2\pi}{q} + o\left(\frac{1}{q}\right). \tag{2} $$

Combining the two, your LHS is asymptotically equivalent to $$ \left\lvert \frac{e^{2\pi i a q^r} -1}{e^{2i\pi\frac{a}{q}}-1}\right\rvert\operatorname*{\sim}_{q\to\infty}\frac{q}{\pi} $$ so for any $\theta < 1$ and large enough $q$ it cannot be upper bounded by $q^\theta$.