Yashar Mermarian writes [here][1] that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that 
$$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$
It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. 
Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. I hope someone can make this clear.

To summarize this gives the following version of a Bursuk Ulam type result:

> If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.



![two semispheres][2]  $U_{\pi/2}$ for two points which are not antipodal. 

  [1]: http://arxiv.org/abs/0911.3972
  [2]: https://i.sstatic.net/5dr8Y.jpg