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).
CORRECTION: You have to allow the "trivial" free factor system in order for this meet to be well-defined, because it is possible that the definition above produces the emptyset. In which case I should have left out the condition "$A \cap A' \ne 1$" in my definition of the meet.
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 $\mathcal{F}'$.