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.