I emailed David Conlon about this question. He agreed to let me share his answer. In short, the problem very much seems to be open (I've added the relevant tag). As Thomas mentions, the upper bound I cite is straightforward. And nothing better is known!
If one looks for papers on Ramsey multiplicity, a few come up, but they deal with a different concept, that I explain below. The quotes are from Conlon's emails.
Unfortunately, the concept you're talking about is also known as the Ramsey
multiplicity! There are very few references as far as I know. The only one I
can think of offhand is the Piwakowski and Radziszowski paper which you
quoted. Perhaps there's something in the references to that paper, but I
doubt it somehow.
Indeed, in the papers I have seen (included P-R, where $m(4)=9$ is proved), there are no arguments about $m(n)$ for general $n$ (or even $n=5$).
The function you're interested in is rather amorphous, I'm afraid. My result
will imply that if $n \ge 4^t$ you must have at least $n^t/2^{3t^2/2}$ copies of
$K_t$ or thereabouts. But when your number is below $4^t$ it implies nothing.
In general, because we don't understand the Ramsey function, I find it hard
to imagine how we might be able to say anything at all about $m(n)$. Unless
there's an elementary argument which gives something. It reminds me of
estimating the difference between successive Ramsey numbers like $r(n,n)$ and
$r(n,n+1)$, where, though the difference is almost certainly exponential, the
largest difference that can be guaranteed is tiny (I think linear or
quadratic even, though I can't remember exactly).
Here, $r(m,n)$ are the usual Ramsey numbers (what I called $R(n)$ in the question, is $r(n,n)$ in this notation). In general, $r(m,n)$ is the smallest $k$ such that any coloring of the edges of $K_k$ with blue and red either contains a blue copy of $K_m$ or a red copy of $K_n$.
The only thing that appears clear to me is an upper bound following from the
probabilistic method, namely $\displaystyle \frac{\binom{r(n)}{n}}{2^{\binom n2}}$. It's not even
obvious how one would approach showing that the multiplicity is at least 2!
Finally, as to the question of how to call this concept:
I'd suggest that this be called the critical multiplicity or something like
that, just to distinguish it from the usual multiplicity function.
The usual Ramsey multiplicity is defined as follows. It is significantly better understood than $m(n)$.
Let $k_t(n)$ be the minimum number of monochromatic copies of $K_t$ within a two-coloring of the
edges of $K_n$, and let $$ c_t(n)= \frac{k_t(n)}{\binom nt} $$ be the minimum proportion of monochromatic copies of $K_t$ in such a two-coloring.
It is known that the numbers $c_t(n)$ increase with $n$. The Ramsey multiplicity of $t$ (or of $K_t$) is $\displaystyle c_t\lim_{n\to\infty} c_t(n)$.
(Relevant references can be found in Conlon's paper mentioned in the question.)
Best Answer
One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.
In http://nucularpower.com/papers/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle; I have been unable to find a 2-coloring that does so. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.
The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.
Edit, December 2016: I have updated the link to my note. In case the link becomes outdated again, the 3-coloring that avoids a degenerate triangle is just the coloring shown below, where the upper hexagon illustrates how boundaries are colored.