Suppose we have a gaussian beam in a medium with $n>1$ with a waist $w_0$ centered at $z = 0$. After a distance of $d_{med}$ the beam leaves the medium and travels through air. To compute the change of the complex beam parameter $q(z)$ one can apply the appropriate ABCD-matrices.
Inside the medium $q(z) = z + i\ z_R$ with $z_R=\frac{\pi w_0^2 n}{\lambda_{vac}}$. The matrix to describe the beam propagation consists of:
$(1)$ propagation in medium for $d_{med}$: $m_1 = \left( \begin{array}{cc} 1 & d_{med} \\ 0 & 1 \\ \end{array} \right) $
$(2)$ refraction at medium/air boundary: $m_2 = \left( \begin{array}{cc} 1 & 0 \\ 0 & n \\ \end{array} \right) $
$(3)$ propagation in air for $d_{air}$: $m_3 = \left( \begin{array}{cc} 1 & d_{air} \\ 0 & 1 \\ \end{array} \right) $
Thus $M = m_3 \times m_2 \times m_1 =\left( \begin{array}{cc} 1 & d_{med} + n \ d_{air} \\ 0 & 1 \\ \end{array} \right)$
and from this $A = 1,\ B = d_{med} + n d_{air},\ C = 0$ and $D = n$.
So one ends up with $q=\frac{A q_0 + B}{C q_0 +D} = \frac{q_0 + d_{med} + n d_{air}}{n}$ with $q_0 = i\ z_R$.
If now $d_{air}\rightarrow 0$, i.e. one approaches the boundary from the right, q is given given by $ \frac{q_0 + d_{med}}{n}$, in contrast to $q=q_0 + d_{med}$ when approaching from within the medium.
Naturally I would say one also needs to change $q_0$ and $d_{med}$, i.e. $d_{med}\rightarrow n\ d_{med}$ and $q_0\rightarrow n\ q_0$ since $\lambda = \lambda_{vac} /n \rightarrow \lambda_{vac}$, but I think of $q_0$ as some constant, which is determined by the mediums properties. So I am missing a physical reason for this operation.
And if this is not the case, although I could live with a discontinuity in $R(z)$, I am troubled with the idea of a sudden jump in the beam radius $w(z)$.
