The question is not stated precisely. So I'm free to say anything :)
If you are interested in "non-uniqueness" then the anser is "NO". In any such triangle $[x y z]$ there are at least two distinct geodesics between directions of $[x y]$ and $[x z]$; thus the space of directions at $x$ can not be a sphere.
If you are interested in "the third side is wrong", the answer is "YES". Take $x=(0,0,0)$, $y=(1,0,1)$ and $z=(0,1,-1)$ in $\mathbb R^2\times [-1,1]$. Then glue $\mathbb R^2\times \{1\}$ to it-self along reflection $(u,v)\mapsto (-u,v)$ and glue $\mathbb R^2\times \{-1\}$ to it-self along reflection $(u,v)\mapsto (u,-v)$. You get a singular Alexandrov space with triangle $[x y z]$ where the side $[y z]$ might be wrong for some filling of the hinge at $x$. BUT this triangle admins a flat filling. It is easy to smooth this spaces near both singular lines to obtain Riemannian manifold with the same property.
I know one example of a triangle in singular Alexandrov 3-space such that all distances between points on sides are the same as the corresponding distances in the model triangle, but it can not be filled with a flat triangle. This is a bit tricky to construct. I can not make this example to be Riemannian (and I feel that it is impossible).
P.S. The answer to the "specific" question is "NO".
Take a 2-dimensional nonnegatively curved manifold $M$ with a pair of points $p,q$ such that there are two minimizing geodesics from $p$ to $q$. One can choose a triangle with vertexes $x=(0,p)$, $y=(1,q)$ and $z=(-1,q)$ in the product $\mathbb R\times M$ which does not admit flat filling. On the other hand any such triangle satisfies your condition for midpoint of $[y z]$.