In Alice/Bob/Cindy terms (EDIT: and with a little more detail):
Alice and Bob have each securely obtained one key of an RSA keypair from a trusted third party. Alice has one key ($e$ and $n$), Bob has the other ($d$ and $n$, where $d\equiv e^{-1} (mod\ \phi(n))$).
The RSA algorithm technically does not care which of the two keys is used to encrypt and which to decrypt; if $c_1 \equiv m^e (mod\ n)$ and $m \equiv c_1^d (mod\ n)$, then $c_2 \equiv m^e (mod\ n)$ and $m \equiv c_2^d (mod\ n)$. ($c_1 \neq c_2$)
Therefore, Alice uses $e$ and $n$ to encrypt a message $m_a$ into ciphertext $c_a$ and sends it to Bob, who decrypts it with $d$ and $n$, then encrypts his response $m_b$ with the same $d$ and $n$, and sends the ciphertext $c_b$ to Alice who can decrypt it with $e$ and $n$. Thus, one pair of asymmetric keys is being used to form a two-way communication channel, instead of the normal one-way usage.
Now, Cindy does not know $e$, $d$, or $n$, or any artifact used to produce them, such as $p$ and $q$. she can only see $c_a$ and $c_b$ as they are passed between Alice and Bob. Given that all other security concerns with RSA are properly handled, such as $\geq$2048-bit keys and OEAP padding scheme, is there an "efficient" way (an attack) by which Cindy can obtain $m_a$ and/or $m_b$?