You can show $\frac{1}{x^q}$ converges on$ (0,1] $ for $q<1$ and that's a bound for the $\frac{|\sin(x)|}{x^q}$ so we know for $q<1$ our function is integrable- I can't seem to improve on this because the answer is $<2$. I decided to use the substitution $u=\frac{1}{x}$ to slay this problem by making the limits $1$ and $\infty$- and using a summation trick with integrals to get an infinite sum of powers of $k$. I would type in latex but it would be horrendously tiring and boring- You get $p\displaystyle\sum_{k=1}^{n-1} k^{q-2} $ (where $p$ is a constant) which looked promising because of the $2$ in the power but we need $q-2<-1$ for this to converge as $n$ tends to $\infty$ i.e sadly $q<1$ again :(. Who can save me please?
P.S I treated the new integral after substitution: $\displaystyle\int_{1}^ {\infty} \sin(\frac{1}{u})u^{q-2}$ as being from $\pi$ to $infinity$ instead of from $1$. That wouldn't make a difference to the problem because it behaves fine between $1$ and $\pi$- and the summation involved making the upper limit $n\pi$ and letting that $n$ tend to $\infty$ if you see where I was going with this.
$\sin x$ behaves like $x$ in a right neighbourhood of the origin, and we just need that the singularity there is integrable, i.e. $p<2$. In such a case: $$ \int_{0}^{1}\frac{\sin x}{x^p}\,dx \leq \int_{0}^{1}x^{1-p}\,dx =\frac{1}{2-p}.$$ The exact value of the integral can be computed by expanding $\sin x$ as a Taylor series: $$ \int_{0}^{1}\frac{\sin x}{x^p}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!(2n+2-p)}.$$