Place the vertices of the equilateral triangle at the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, and denote the radius of the semicircles by $r$. The equilateral triangle formed by the diameters of the semicircles has side length $2\sqrt2$, so its area is $2\sqrt3$. To find the distance from a vertex where the diameter corresponding to that vertex intersects the semicircle corresponding to another vertex, take the diameter
$$
\pmatrix{1\\0\\0}+\lambda\pmatrix{0\\1\\-1}
$$
and find the point on it at distance $r$ from the vertex $\pmatrix{0\\0\\1}$:
$$
1+\lambda^2+(1+\lambda)^2=r^2\;,
\\
2\lambda^2+2\lambda+2=r^2\;,
\\
\lambda=-\frac12+\sqrt{\frac{r^2}2-\frac34}\;.
$$
Thus the side length of the three smaller equilateral triangles that are "missing" is
$$
\sqrt2\left(1-\left(-\frac12+\sqrt{\frac{r^2}2-\frac34}\right)\right)=\sqrt2\left(\frac32-\sqrt{\frac{r^2}2-\frac34}\right)\;,
$$
so their total area is
$$
3\cdot\frac{\sqrt3}4\cdot2\left(\frac94+\frac{r^2}2-\frac34-3\sqrt{\frac{r^2}2-\frac34}\right)=\frac34\sqrt3\left(3+r^2-3\sqrt{2r^2-3}\right)\;.
$$
Subtracting this from $2\sqrt3$ leaves
$$
\frac{\sqrt3}4\left(9\sqrt{2r^2-3}-3r^2-1\right)\;.
$$
Now we need to add back the three circular segments that extend into the three equilateral triangles we subtracted. Their radius is $r$, and their angle $\alpha$ is twice the angle between $\pmatrix{1\\\lambda\\-\lambda}-\pmatrix{0\\0\\1}$ and $\pmatrix{1\\1\\-1}-\pmatrix{0\\0\\1}$ and thus
$$
\alpha=2\arccos\frac{3(\lambda+1)}{\sqrt{6\cdot(2\lambda^2+2\lambda+2)}}=2\arccos\left(\sqrt{\frac38}\frac{1+\sqrt{2r^2-3}}r\right)\;.
$$
With the formula for the area of a circular segment, the total area is then
$$
\frac{\sqrt3}4\left(9\sqrt{2r^2-3}-3r^2-1\right)+\frac32r^2\left(\alpha-\sin\alpha\right)\;.
$$
Here's a plot of this total area for $r$ in the relevant range $[\sqrt2,\sqrt6]$, with the area ranging from $\pi-\sqrt3$ to $2\sqrt3$.
In this calculation the side length of the equilateral triangle was fixed at $\sqrt2$; for a different side length $a$, scale the radius by $\sqrt2/a$ and then scale the resulting area by $a^2/2$.
