There seems to be some confusion here regarding language. In your first paragraph, you define the free monad generated by an endofunctor $R$, i.e. one first fixes $R$ then gets the monad. A good toy example for such $R$ would be $U\circ F$ where $F:X\to Y$ is the free functor into some category $Y$ which is well understood and $U:Y\to X$ is the forgetful functor. So the answer to your first question is yes, it can take this form and that's a nice case to play with. But in general $R$ need not take this form. Theorem 5.5 in the paper you cite gives a sufficient condition on an endofunctor $R$ so that free monad on $R$ exists. That condition has nothing to do with $R$ coming from an adjunction.
I can interpret your second question in a couple of ways. One way is as "when is $T$ of the form $U\circ F$?'' This is trivial; it's well-known that a monad $T$ is always of the form $U\circ F$ (see wikipedia), often for more than one choice of $U,F$. A better way to interpret your question is "given a monad $T$, how do I determine if it's the free monad generated by some $R$ which takes the form $U\circ F$?" This appears to be a non-trivial problem. Theorem 5.4 in the paper gives one situation where you can determine that your monad is free and recover $R$, but this doesn't classify all such situations. Anyway, even in this nice case where you get $R$ in hand, there are plenty of times such $R$ could fail to be in the form $U\circ F$. For instance, this MO answer shows that any such $R$ would need to be a homotopy equivalence on the nerve of $X$. It should not be too hard to construct an endofunctor for which this fails, since there are plenty of self-maps of simplicial complexes which are not homotopy equivalences.
