To translate the graph of $y= \sqrt{2}\sin(2x+ \frac{\pi}{3})$ to the right by $\varphi$ ($0<\varphi<\pi$) units, we get the graph of $y= \sqrt{2}\sin(2x-2\varphi+ \frac{\pi}{3})$.
According to the problem, we have $y=2\sin x(\sin x-\cos x)-1=2\sin^2x-\sin 2x-1=-\sin 2x-\cos 2x=- \sqrt{2}\sin(2x+ \frac{\pi}{4})= \sqrt{2}\sin(2x+ \frac{5\pi}{4})$.
Therefore, $-2\varphi+ \frac{\pi}{3}= \frac{5\pi}{4}+2k\pi$, where $k\in \mathbb{Z}$. This implies $\varphi=-k\pi- \frac{11\pi}{24}$. Thus, $\varphi= \boxed{\frac{13\pi}{24}}$.
This problem involves using the transformation rules of the graph of the function $y=A\sin(\omega x+\varphi)$ and deriving formulas from these rules.
The main focus of this problem is on the application of the transformation rules and formula derivation for the function $y=A\sin(\omega x+\varphi)$, which is considered a basic problem.