**This is not true for $S^2$.** Namely, one can construct two convex non-isometric spherical triangles such that all medians have length $\frac{3\pi}{4}-\varepsilon$.

1) The first triangle is the standard equilateral triangle. Clearly medians of such triangles all have the same length and their length vary from $0$ to $\pi$.

2) The second triangle will be a perturbation of the following one. Take on $S^2$ two opposite points $A$ and $B$ and join them by two geodesics (of length $\pi$) that bound a sector with angle $\frac{3\pi}{4}$. Call one of these geodesics $AB$. And choose $C$ as the mid point of the other geodesic. It is easy to see that for this degenerate triangle all medians have length $\frac{3\pi}{4}$. Now, by perturbing slightly $A$, $B$ and $C$ one can decrease the lengths of all medians by the same amount.

*Note.* It is important in the second construction that by varying $A$ and $B$ as little as we want, we can achieve that the median starting from $C$ takes any length in $(0,\pi)$. However the length of two other medians stay close to $\frac{3\pi}{4}$.

**PS, hyperbolic case**. Concerning the hyperbolic case, it looks like the answer is positive.
If I would like to prove it, I would do as follows.

i) Consider the map from the convex cone in $\mathbb R_{\ge 0}^3$ consisting of triples $(a,b,c)$ satisfying the (non-strict) triangle inequality to itself: lengths of sides $\to$ lengths of medians.

ii) Realise that this map if proper (preimage of compact is compact). And it is an isomorphism on the boundary on the cone.

iii) The map is a "diffeo" close to the point $(0,0,0)$ - because it is so for Euclidean triangles.

**iv)** *This is the most complicated bit* -- show that the map has non-vanishing differential. This is where one needs to make a calculation. But it looks very plausible.

v) If all the above holds then the map is an isomorphism. QED