Although HW has answered your question negatively for greatest lower bounds, there is a closely related poset for which greatest lower bounds and least upper bounds indeed exist. Instead of an individual free factor $A$, first consider its conjugacy class $[A]$. Then, instead of individual conjugacy classes of free factors $[A]$, consider a "free factor system" (in the language of Bestvina-Feighn-Handel): a finite set $\mathcal{F} = \{[A_1],\ldots,[A_k]\}$ such that there exists a free factorization of the form $F_n = A_1 * \cdots * A_k * B$ where $B$ may or may not be trivial but all the $A_i$'s are nontrivial. The partial ordering $\mathcal{F}\sqsubset \mathcal{F}'$ is defined by requiring that for each $[A] \in \mathcal{F}$ there exists $[A'] \in \mathcal{F}'$ such that $A$ is conjugate to a subgroup of $A'$.
The Kurosh Subgroup Theorem can be translated into the statement that this poset has greatest lower bounds. The greatest lower bound of $\mathcal{F}$ and $\mathcal{F}'$ is $$\mathcal{F} "meet" \mathcal{F}' = \{[A \cap A'] \, | \, [A] \in \mathcal{F}, [A'] \in \mathcal{F}', A \cap A' \ne 1\} $$ (I don't know how to get the "meet" operator in this version of TeX).
The meet operator can then be extended to an operator on arbitrary sets of free factor systems, and using this one gets least upper bounds too: the least upper bound of $\mathcal{F}$ and $\mathcal{F}'$ is the meet of all free factor systems $\mathcal{F}''$ (including the improper free factor system $\{[F]\}$) such that $\mathcal{F} \sqsubset \mathcal{F}''$ and $\mathcal{F}' \sqsubset \mathcal{F}''$. This $\mathcal{F}''$ is called the ``free factor support'' of $\mathcal{F}$ and $\mathcal{F}'$.