This is a variation of Angelo's first example. Consider a surface without rational curves and blow-up a finite number of distinct points. If $E$ is one of the exceptional divisors and $f \colon E \to X$ is the inclusion in the blow-up, since
$H^0(E, N_{E|X})=H^1(E, N_{E|X})=0$,
we obtain $H^1(E, f^*T_X)=H^1(E, T_E)=0$, so $X$ is convex. On the other hand, $X$ is not homogeneous, since its automorphism group does not act transitively.
ADDED. As remarked by mdeland, this example does not really work since we can take finite covers of $E$ in order to obtain curves that violate the convexity condition. In order to avoid this problem, we must consider examples such that the splitting type of the tangent bundle of $X$ over the rational curve does not contain summands of negative degree. One such example is the following.
Let $C$ be any curve of genus $g(C) \geq 1$ and let $\mathcal{E} = \mathcal{O}_C \oplus \mathcal{L}$, where $\mathcal{L}$ is a line bundle of negative degree $-e$. Then $X = \mathbb{P}(\mathcal{E}) $ is a ruled surface over $C$ which contains a unique section $C_0$ such that $C_0^2 = -e$, in particular $X$ is not homogeneous.
Now let us show that $X$ is convex. Let $F \cong \mathbb{P}^1$ be any fibre of $p \colon X \to C$. We have a short exact sequence
$0 \to T_F \to (p^*T_X)|_F \to N_F \to 0$.
Since $T_F=\mathcal{O}_{\mathbb{P^1}}(2)$ and $N_F=\mathcal{O}_F$, it follows $\textrm{Ext}^1(N_F, T_F)=0$. Therefore the sequence above actually splits and we obtain
$(p^*T_X)|_F = \mathcal{O}(2) \oplus \mathcal{O}$.
On the other hand, since $g(C) \geq $ the unique rational curves on $X$ are the fibres of $p$, so every holomorphic map $f \colon \mathbb{P}^1 \to X$ is given by the inclusion of a fibre composed with a finite cover. If $d$ is the degree of such a cover, we obtain
$(f^*T_X)|_F=\mathcal{O}(2d) \oplus \mathcal{O}$.
It follows $H^1(\mathbb{P}^1, f^*T_X)=0$, so $X$ is convex. Notice that $X$ is uniruled, but not rationally connected.