The second fundamental form $II$ of $\partial V$ vanishes in the radial direction (along any line that goes through the origin), so one eigenvalue of $II$ is zero.
There are only two eigenvalues since $\dim(\partial V)=2$ and the mean curvature is (up to a factor of 2) the trace of $II$.
Therefore the nonzero eigenvalue of $II$ has the same sign as the mean curvature.

We have thus that $II$ is positive semidefinite iff the mean curvature is positive (and similarly for negative).
Positive semidefiniteness of $II$ is equivalent with convexity, so the answer to your question is yes.

Be careful with the last assertion, though.
For a submanifold on a general Riemannian manifold positive semidefiniteness of $II$ is not enough to guarantee convexity in the geodesic sense.
Convexity does always imply $II\geq0$.

Let's see why things work in our case.
Let $\gamma:[a,b]\to\mathbb R^3\setminus\{0\}$ be any line segment, and suppose $\gamma(a),\gamma(b)\in V$.
Let $\phi:\mathbb R^3\setminus\{0\}\to S^2$ be the projection to the sphere, $\phi(x)=x/|x|$.
Now $\phi\circ\gamma:[a,b]\to S^2$ is an arc of a great circle on the sphere.
The second fundamental form on $\partial V$ is positive semidefinite and vanishes in the radial direction, so $II$ on $\partial\phi(V)$ is also positive semidefinite.
It is an easier exercise to see that this implies that $\phi(V)$ is convex (in the sense that a minimal geodesic joining any two points in the set stays in the set).
Thus $\phi\circ\gamma$ only takes values in $\phi(V)$.
Since $V$ is conical, this means that $\gamma$ only takes values in $V$.