Yes, an abelian variety $A$ contains no rational curves. If $C$ is a curve on an abelian variety then $C$ has regular $1$-forms coming from $1$-forms on $A$.

Yes, an abelian variety $A$ contains no rational curves.

Suppose not and let $f: \mathbb P^1 \to A$ be a non-constant morphism.

If $f$ is inseparable then it must be the composition of some power of Frobenius of $\mathbb P^1$ with a non-constant separable map $g: \mathbb P^1 \to A$. Thus we may assume that $f$ is separable, i.e., $df : T \mathbb P^1 \to f^{\ast} T A$ is not the zero morphism. Therefore the general $1$-form $\omega \in H^0(A,\Omega^1)$ will give rise to a non-zero $1$-form $f^{\ast} \omega$ on $\mathbb P^1$. Contradiction.

Remarks:

1. Above, I have expanded the original answer "If $C$ is a curve on an abelian variety then $C$ has regular $1$-forms coming from $1$-forms on $A$" in order to incorporate Voloch's comment about positive characteristic.
2. The same argument shows that non-algebraic compact complex tori contain no rational curves.
3. Since over $\mathbb C$ the Albanese variety of a compact Kahler manifold $X$ is usually defined as $H^0(X, Omega^1)^{\ast} / H_1(X, \mathbb Z)$, the argument above is essentially the same as Voloch's when the characteristic zero.
4. Let $X$ be a smooth projective variety and $f:X \to A$ be a morphism. If $df$ has maximal rank then $H^0(X,{\Omega^i}) \neq 0$ for every $i \le \dim X$. Thus an abelian variety contains no subvarieties without regular forms in any particular degree.