Solution:
(I) The function $f(x)=\sin (\omega x- \frac {\pi}{6})+\sin (\omega x- \frac {\pi}{2})$
$=\sin \omega x\cos \frac {\pi}{6}-\cos \omega x\sin \frac {\pi}{6}-\sin ( \frac {\pi}{2}-\omega x)$
$= \frac { \sqrt {3}}{2}\sin \omega x- \frac {1}{2}\cos \omega x$
$= \sqrt {3}\sin (\omega x- \frac {\pi}{3})$,
Also, $f( \frac {\pi}{6})= \sqrt {3}\sin ( \frac {\pi}{6}\omega- \frac {\pi}{3})=0$,
$\therefore \frac {\pi}{6}\omega- \frac {\pi}{3}=k\pi$, $k\in\mathbb{Z}$,
Solving gives $\omega=6k+2$,
And since $0 < \omega < 3$,
$\therefore \boxed{\omega=2}$;
(II) From (I), we know that $f(x)= \sqrt {3}\sin (2x- \frac {\pi}{3})$,
Stretching the x-coordinates of the points on the graph of the function $y=f(x)$ by a factor of $2$ (keeping the y-coordinates unchanged), we get the graph of the function $y= \sqrt {3}\sin (x- \frac {\pi}{3})$;
Then shifting this graph to the left by $\frac {\pi}{4}$ units, we get $y= \sqrt {3}\sin (x+ \frac {\pi}{4}- \frac {\pi}{3})$,
$\therefore$ the function $y=g(x)= \sqrt {3}\sin (x- \frac {\pi}{12})$;
When $x\in[- \frac {\pi}{4}, \frac {3\pi}{4}]$, $x- \frac {\pi}{12}\in[- \frac {\pi}{3}, \frac {2\pi}{3}]$,
$\therefore\sin (x- \frac {\pi}{12})\in[- \frac { \sqrt {3}}{2},1]$,
$\therefore$ when $x=- \frac {\pi}{4}$, $g(x)$ reaches its minimum value of $- \frac { \sqrt {3}}{2}\times \sqrt {3}=\boxed{- \frac {3}{2}}$.