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 $a^2/2$.
This is supposed to be a comment but I would like to post a picture.
For any $m \ge 3$, we can put $m+2$ vertices on the unit sphere
$$( 0, 0, \pm 1) \quad\text{ and }\quad \left( \cos\frac{2\pi k}{m}, \sin\frac{2\pi k}{m}, 0 \right) \quad\text{ for }\quad 0 \le k < m$$
Their convex hull will be a $m$-gonal bipyramid which appear below.
Up to my knowledge, the largest $n$-vertex polyhedron inside a sphere is known only up to $n = 8$.
- $n = 4$, a tetrahedron.
- $n = 5$, a triangular bipyramid.
- $n = 6$, a octahedron = a square bipyramid
- $n = 7$, a pentagonal bipyramid.
- $n = 8$, it is neither the cube ( volume: $\frac{8}{3\sqrt{3}} \approx 1.53960$ ) nor the hexagonal bipyramid ( volume: $\sqrt{3} \approx 1.73205$ ). Instead, it has volume
$\sqrt{\frac{475+29\sqrt{145}}{250}} \approx 1.815716104224$.
Let $\phi = \cos^{-1}\sqrt{\frac{15+\sqrt{145}}{40}}$, one possible set of vertices are given below:
$$
( \pm \sin3\phi, 0, +\cos3\phi ),\;\; ( \pm\sin\phi, 0,+\cos\phi ),\\
(0, \pm\sin3\phi, -\cos3\phi),\;\; ( 0, \pm\sin\phi, -\cos\phi).
$$
For this set of vertices, the polyhedron is the convex hull of two polylines.
One in $xz$-plane and the other in $yz$-plane. Following is a figure of this polyhedron,
the red/green/blue arrows are the $x/y/z$-axes respectively.
$\hspace0.75in$ 
For $n \le 8$, above configurations are known to be optimal. A proof can be found
in the paper
Joel D. Berman, Kit Hanes, Volumes of polyhedra inscribed in the unit sphere in $E^3$
Mathematische Annalen 1970, Volume 188, Issue 1, pp 78-84
An online copy of the paper is viewable at here (you need to scroll to image 84/page 78 at first visit).
For $n \le 130$, a good source of close to optimal configurations can be found
under N.J.A. Sloane's web page on
Maximal Volume Spherical Codes.
It contains the best known configuration at least up to year 1994. For example,
you can find an alternate set of coordinates for the $n = 8$ case from the maxvol3.8
files under the link to library of 3-d arrangements there.
Best Answer
I think it is the circle by the following reasoning. I'm not totally sure about this though, so if there is an error hopefully someone can correct me.
The shape of maximal area must be bounded by a curve of constant width (with the diameter being the width), since any curve of varying width could be expanded in the directions where the width is deficient, increasing the area while keeping the diameter the same.
Now by Barbier's theorem, the perimeter of any curve of constant width $w$ is just $\pi w$, regardless of the shape. Thus all shapes that can be candidates for optimality must have the same perimeter.
But by the isoperimetric inequality, a circle is the curve enclosing maximal area among all shapes with a given perimeter. Thus the circle encloses the maximal area among all shapes with a given diameter.