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I would guess that an $\mathbb R$-monoid is the same thing (once you put extra smoothness axioms on the $\mathbb R$-action) as a Lie group whose exponential map $\mathfrak g\to G$ is a diffeomorphism. The Heisenberg group $$ \left(\begin{matrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \\ \end{matrix}\right) $$ is the simplest non-commutative example of such a group. Take the same example but restrict $*\in \mathbb Q$ to get an example of a $\mathbb Q$-monoid which is not an $\mathbb R$-monoid.