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.


 - 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).