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In addition to the given diagram, we can draw lines $\overline{SR}$ and $\overline{RT}.$ The area of rhombus $SRTU$ is half the product of its diagonals, which is $\frac{2\sqrt{3}\cdot 2}{2}=2\sqrt{3}$. However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by $\frac{1}{6}$, then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is $2(\frac{4\pi }{6}-\sqrt{3}).$ The area of rhombus $SRTU$ minus the circular segments is $2\sqrt{3}-\frac{4\pi }{3}+2\sqrt{3}=\boxed{4\sqrt{3}-\frac{4\pi }{3}}.$