In the online book, page 312, http://projecteuclid.org/download/pdf_1/euclid.pl/1235419485
what cardinal notion do we get, by requireing that X in the bottom of the page, is not only stationary (yielding ineffable) or unbounded (weakly compact), but even club.
You get the very famous large cardinal called inconsistency.
Theorem: If any $f\colon [\kappa]^2\to 2$ has a homogeneous club then the club filter on $\kappa$ is an ultrafilter.
Proof: By a standard argument we also have the property for colourings with 3 colours. Now take an arbitrary $X\subseteq\kappa$ and define $f\colon [\kappa]^2\to 3$ by $$f(x,y)=\begin{cases}0;& x,y\in X\\ 1;& x,y\notin X\\ 2;&\text{otherwise}\end{cases}$$
Let $C$ be the homogeneous club for $f$. Clearly $C$ cannot have colour $2$ (since from two pairs, both of which have exactly one element of $X$, we can get a pair with both elements in $X$). But if $C$ has colour $0$ then $C\subseteq X$, putting $X$ into the club filter, and if $C$ has colour $1$ then $C\cap X=\emptyset$, putting $\kappa\setminus X$ into the club filter. $\square$
But since there are stationary costationary subsets of $\kappa$ (at least assuming choice), the club filter cannot possibly be an ultrafilter.
Edit: Here is a better argument showing that your proposed partition property is inconsistent.
Proof: The partition property clearly implies that any function $f\colon \kappa\to 2$ is constant on a club. Now apply this fact to the characteristic function of a stationary costationary subset of $\kappa$ to get a contradiction. $\square$