I don't have a proof, but I have an idea which suggests the answer is no, the minimal ellipse may not be unique. Right or wrong, I hope someone will generate a picture to illustrate the idea, and see if in addition there is a near-octagon which exhibits the pair of minimal ellipses.
I use eight-fold symmetry and restrict attention to the positive quadrant. Draw a quarter circle of unit radius, a nearly circular ellipse quarter with major axis $1+ \epsilon$ and minor axis $1/(1+\epsilon)$, and a reflection about $x=y$ of this quarter ellipse. (The ellipses axes lie on the axes bordering the quadrant.) Consider the point of intersection $P$ between the ellipses and the line x=y. One vertex of the proposed octagon will be $P$, and another will be the point $Q=(0, 1+ \epsilon)$. There will be a curve between the two which bisects that portion of the symmetric difference between the two ellipses.
I challenge the illustrator to find a curve which does such an area bisection, and induces a near octagon which does not have a smaller symmetric difference with the unit circle. I believe it possible because 1) the point $P$ is far enough in the interior of the circle that much of the circle "sticks out", and 2) the freedom one has in bisecting the portion of the ellipse symmetric difference. This may not prove that the given ellipses are minimal, but it may be possible to draw the curve to show that any minimal ellipse must have a reflection which is also minimal.
Gerhard "Easily Writes One Thousand Words..." Paseman, 2016.12.23
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[![OctGP][1]][1]
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<sup>Figure added by J.O'Rourke. $\epsilon=\frac{1}{10}$.</sup>
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[1]: https://i.sstatic.net/bXWmQ.jpg