Given some joint state $\rho_{AB}$, one can find either the marginal state $\rho_A$ or the marginal state $\rho_B$ through a CPTP map. The proof being that partial tracing is indeed CPTP.
Is a CPTP map that outputs $\rho_A\otimes\rho_B$ possible?
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No, because such a map would be non-linear. You want to perform the operation $$ \rho_{AB} \mapsto \operatorname{tr}_B(\rho_{AB}) \otimes \operatorname{tr}_A(\rho_{AB}). $$
Take for example $\rho_{AB} = \lambda \phi^+ + (1-\lambda)|00\rangle\langle 00|$. If the map were linear, you could apply it to each term individually, and obtain $\lambda I/4 + (1-\lambda)|00\rangle\langle 00|$. But if you apply it to the state as a whole, what you obtain is $$\Big(\lambda I/2 + (1-\lambda)|0\rangle\langle0|\Big)\otimes \Big(\lambda I/2 + (1-\lambda)|0\rangle\langle0|\Big).$$
