If every divergent curve has unbounded lenght then the manifold is complete.

Every divergent curve has unbounded length $\implies$ Completeness:

Suppose the manifold was not complete. Then, there exists a geodesic $\gamma:[0,T) \to M $ which cannot be extended further than $T$. It is clear that this curve has bounded length (specifically, $k.T$ for some constant $k$). However, $\gamma$ is divergent. This is a standard result of maximal solutions of ODEs, but we can make a particular simple proof here: Suppose it was not divergent. Then it is contained in some compact set. Consider the sequence $\gamma(t_n)$, with $t_n \to T$. It has a convergent subsequence $\gamma(t_{n_i})\to p$. Consider this point $p$, and a strongly normal neighbourhood chart domain(*) around it. Now, take a point $\gamma(\xi)$ ($\xi$ sufficiently close to $T$) such that it is in this chart. Since $\gamma(t_{n_i}) \to p$, the line segment (in the local chart on $T_{\gamma(\xi)}M$) emanating from $\gamma(\xi)$ must pass through $p$. By passing a little more through $p$ with the segment (which is possible because we are in an open set), we would be able to extend the geodesic, which is a contradiction.

(*) Strongly normal in the sense that it is a normal neighbourhood of each one of its points.

Completeness $\implies$ Every divergent curve has unbounded length:

Note that if the manifold is complete, Hopf-Rinow holds. Therefore, the closed ball of radius $N$ ($N$ a natural number) is compact. Your hypothesis says that the curve must leave this ball. Note that this holds for any $N$. What does this imply?

By definition of the metric on a Riemannian manifold, this implies that, for a divergent curve $\alpha$, the $t_0$ given by the definition of divergence (when considering the compact being the closed ball of radius $N$) is such that the length of $\alpha|_{[0,t_0]}$ is greater than $N$. Therefore, $$\int_0^{\infty} \Vert \alpha '\Vert \geq \int_0^{t_0} \Vert \alpha '\Vert \geq N .$$