This is going to be more a non-answer than an answer:

I don't think there is a useful notion of morphism between Frobenius manifolds, besides the notions of (local) isomorphisms, and of a sub-Frobenius manifold (which is not the same thing as a submanifold of a Frobenius manifolds).

To argue completely in intrinsic terms: if you require that a morphism preserves the metric on the tangent spaces, then it is necessarily a local embedding. Locally, the image is then a sub-Frobenius manifold, i.e. a flat submanifold $M \subset N$ such that the multiplication on the tangent bundle $T_N \otimes T_N \to T_N$ maps $T_M \otimes T_M$ to $T_M$.

Examples of sub-Frobenius manifolds arise naturally, e.g. in quantum cohomology,
$\oplus_p H^{p, p}(X, \mathbb{C})$ is a sub-Frobenius manifold of $H^*(X, \mathbb{C})$.

To argue with a little more context: a more flexible notion of morphisms of Frobenius manifolds wouldn't be useful unless it arises naturally. However, even for the most simple situations you could think of, say if you compare the quantum cohomology of a product $X \times Y$ with the quantum cohomology of $X$ and $Y$, there does not seem to be a useful morphism between the Frobenius manifolds involved. Instead, the (Frob. mfd of) QC of $X \times Y$ is a tensor product of the QCs of $X$ and $Y$. Other nice statements are known for Grassmannians $G(k, n)$, whose QC is a kind of mixture of anti-symmetric and symmetric $k$-fold tensor product of the QC of $P^{n-1}$ (see papers by subsets of Bertram/Ciocan-Fontanine/Kim/Sabbah on abelian/non-abelian quotients). Again, a very nice statement, but no morphisms of Frob. manifolds anywhere.

[If I may sneak in a little advertising: If $\tilde X$ is the blow-up of $X$ at a point, then the QC of then the QC of $\tilde X$ has a partial compactification, and the boundary is a Frob. submanifold isomorphic to QC of $X$. Well, unfortunately that's a little bit of a lie, as the multiplication has a pole on the added boundary divisor, and so arXiv:math.AG/0403260 doesn't read quite as nicely as the statement above. But again, while there is almost an inclusion of QC of $X$ into QC of $\tilde X$, there is no morphism in the other direction.]