I think the short answer is "No", you cannot deduce from this result a classification of all geometric outer automorphisms.
I think it might eventually be possible to obtain a classification of geometric outer automorphisms from the more powerful "relative train track / lamination" machinery developed in the works of Bestvina, Feighn, and Handel, although that has not been done. Nonetheless one can deduce bits and pieces of such a classification.
For instance, suppose that $\phi \in Out(F_n)$ has an attracting lamination $\Lambda$ with the property that the smallest free factor of $F_n$ that supports $\Lambda$ is the whole free group. In this case one can prove that $\phi$ is geometric if and only if there exists a finite $\phi$-invariant set of root-free conjugacy classes $c_1,...,c_k$ such that the smallest free factor of $F_n$ that supports $c_1,...,c_k$ is also the whole free group, $c_1,...,c_k$ are the only root-free conjugacy classes that are not attracted to $\Lambda$ under iteration of $\phi$, and a few other nondegeneracy conditions hold. The picture to keep in mind is that $c_1,...,c_k$ represent the boundary components of a surface on which $\phi$ is represented as a pseudo-Anosov homoemorphism with unstable lamination $\Lambda$. The "nondegeneracy" conditions I mentioned are needed to avoid counterexamples where, say, three of the $c's$ are identified to the same closed curve, and these conditions can be expressed in an intrinsic manner in terms of ``Nielsen theory'' which means the asymptotic behavior of automorphisms representing the outer automorphism $\phi$.
This statement can be found in a slightly different form in Proposition 2.38 of the paper Subgroup classification in $Out(F_n)$ by Handel and myself, and in this exact form in the soon-to-appear Part III of the expanded version "Subgroup decomposition in $Out(F_n)$".