If there are no restrictions on $S$, one can apply the following construction.
Let $$Q(x)=S(x)\prod_{j=1}^r(x-\lambda_j)^2\prod_{j=r+1}^m(x-\lambda_j).$$
We want to construct $S$. Consider first the case when $R$ does not change sign on the real
line. Suppose it is positive. Then choose $S$ so that
a) All double zeros of $Q$ are maxima. This is achieved by inserting zeros of $S$ at properly chosen places.
b) That the minima of $Q$ on each interval between its zeros are less
(more negative) than $-\max R$
(the maximum of $R$ is taken over some big interval that contains all zeros of $Q$).
This can be achieved just by multiplying $S$ on a sufficiently large constant.
Now when you graph $-R$ and $Q$ on the same picture, you see that the graphs intersect as many
times as the degree of $Q$ is. So $R+Q$ has all zeros real. You also make them simple,
by another multiplication of $Q$ on a constant.
The general case is similar. Consider the real zeros of $R$, say $x_1,...,x_k$.
On the intervals $(x_k,x_{k+1})$ where $R(x)>0$, you want all double zeros of $Q$ to be maxima.
On the intervals $(x_k,x_{k+1})$ where $R(x)<0$, you want all double zeros of $Q$ to be minima.
This is achieved by placing the zeros of $S$ at appropriate places. Each new zero of $S$
switches the sign of $Q$ at this place.
Now after you achieved this (that all double roots of $Q$ are maxima where $R$ is positive, and
minima where it is negative), it only remains to multiply $Q$ by a very large positive constant,
to make sure that the graphs of $-R$ and $Q$ intersect as many times as needed.
Finally, by slightly perturbing the constant multiple in $S$, you achieve that all
these real roots are simple.
But I don't now how useful this polynomial can be, as its degree can be larger than the number
of interpolation nodes. You can try to minimize the degree of $S$, the answer will depend
on the relative position of the double roots $\lambda_j$ among all $\lambda_j$.
But roughly speaking the degree of $Q$ constructed with this method, can be about twice as large as that of $R$.