This will be a very sketchy argument, perhaps to be elaborated on later.
First thing we may note is that a short geodesic curve $C$ joining two points $p$ and $q$ is itself a convex set. So if the two metrics are to have the same convex sets, they must also have the same geodesic curves (up to reparametrisation if by geodesic we mean a path of constant velocity). (I'll use curve to mean the set $C$ and path for the parametrised map $\gamma(t)$ to help keep these two apart.)
So now we need to demonstrate that for two metrics, say $g$ and $\tilde g$, to have the same geodesic curves, the two must be related by a linear relation $g_{ij}=a^r_i a^s_j \tilde g_{rs}$ for some constant $a^r_i$.
For a given point, $O$, we can always find a coordinate system which has $O$ at the origin, and so that at $O$ we have $\tilde g_{ij}=\delta_{ij}$ and Christoffel symbols $\tilde\Gamma^k_{ij}=0$ for $\tilde g$. Thus, a geodesic path $\gamma$ (with constant speed) through $O=\gamma(0)$ should have vanishing second derivative at $O$ (in this coordinate system). If $\gamma$ is a path along the geodesic curve (not necessarily with constant speed), instead $\gamma''(0)$ need only be parallel to the direction $\gamma'(0)$.
For the metric $g$, a geodesic path through $O$ should satisfy
$\gamma''_k+\gamma'^i\gamma'^j\Gamma^k_{ij}=0$ at the origin. For this to form a geodesic path under the metric $\tilde g$, we must then for all tangent vectors $u$ have $u^iu^j\Gamma^k_{ij}=\lambda u^k$ for some $\lambda$. In dimensions $2$ and higher, this forces $\Gamma^k_{ij}=0$, which in turn means the derivative $g_{ij,k}=0$ at $O$.
Unless I'm missing something, the rest should be purely technical to show that this requires that $g_{ij}$ and $\tilde g_{rs}$ are related by a constant linear transformation. For flat metrics, any linear transformation should work, while for curved metrics I suspect linear transformations other than scaling will only be possible in "flat directions".
I'll try to address the second question of when a ball of radius $R$ in one metric will remain convex in the other. I'll ignore global issues (not sure if they will matter) and only look at the question locally. This will be even more sketch then the above, and I've probably managed to sneak in a few mistakes on the fly, to read it for ideas only.
Again, to ensure this property locally, a geodesic curve in one metric should have curvature $\kappa$ at most $1/R$ in the other. To verify this, note that a short (infinitesimal) geodesic between points $p,q$ in one metric, in the other metric needs to be inside every ball of radius $R$ with $p,q$ on the surface.
Imagine we have a geodesic curve in metric $\tilde g$ through a point $O$. Also, we select a coordinate system centered around $O$ in which $\tilde g_{ij}=\delta_{ij}$ and $\tilde\Gamma^k_{ij}=0$ at the origin. We then parametrise the curve using the path $\gamma(s)$ (or $\gamma^k$ for coordinate $k$) so that
$\tilde g(\gamma',\gamma')=\tilde g_{ij} {\gamma'}^i {\gamma'}^j=1$ (where $\gamma'=d\gamma/ds$);
and since the curve is geodesic and $\tilde\Gamma^k_{ij}=0$, the second derivative $\gamma''=0$.
In the coordinate system $g$, this curve will have covariant derivative $D\gamma^k={\gamma'}^i{\gamma'}^j\Gamma^k_{ij}$ since we have conveniently selected a coordinate system in which $\gamma''=0$. Subtracting the acceleration part, $D\gamma-\gamma'\cdot g(\gamma',D\gamma)/g(\gamma',\gamma')$, the curve has curvature $\kappa$ where
$$
\kappa^2
=\frac
{|D\gamma-\gamma'\cdot g(\gamma',D\gamma)/g(\gamma',\gamma')|^2_g}
{|\gamma'|^4_g}
$$
which we can then write out in terms of $\gamma'$ and $\Gamma^k_{ij}$. Since we requires $\kappa\le 1/R$ for all such curves, the derived relation becomes
$$
u^iu^ju^lu^m\Gamma^k_{ij}\Gamma^n_{lm}g_{km}
-\frac{\left(u^iu^ju^r\Gamma^k_{ij}g_{kr}\right)^2}{u^pu^qg_{pq}}
\le\frac{\left(u^pu^qg_{pq}\right)^2}{R^2}
$$
for all vectors $u^i$.
So far, things were a bit technical.
Next, although the Christoffel symbols $\Gamma^k_{ij}$ depend on the choice of coordinate system, the difference $\Delta\Gamma^k_{ij}=\Gamma^k_{ij}-\tilde\Gamma^k_{ij}$ between the two metrics is independent of coordinate system changes and thus transform like tensors. The reason for this is that under a change in coordinates from $x^i$ to $\hat x^a$, they transform as
$$
\hat\Gamma^c_{ab}x^k_{,c}=x^i_{,a}x^j_{,b}\Gamma^k_{ij}+x^k_{,ij}
$$
where the second part of the sum is the same for both metrics.
Since the original bound was expressed in a coordinate system with $\tilde\Gamma=0$, we can just plug in $\Delta\Gamma$ instead. To also show that geodesic curves in $g$ have curvature at most $1/R$ in $\tilde g$, all we need do is to swap the two metrics: in the inequality, $\Delta\Gamma$ changes sign (no effect there since it is squared), and $g$ must be replaced by $\tilde g$.
