Fred Galvin showed that if $c:[\mathbb{R}]^2\to\lbrace0,1\rbrace$ is such that $c^{-1}(0)$ and $c^{-1}(1)$ both have the Baire property, then there is a perfect set $P \subseteq \mathbb{R}$ which is $c$-homogeneous. (Note that perfect sets have size $2^{\aleph_0}$.)
Since Borel sets have the Baire property and perfect sets are Borel, Galvin's Theorem answers your second question.
Shelah has shown that it is relatively consistent with ZF+DC that every subset of any Polish space (like $[\mathbb{R}]^2$) has the Baire property. Galvin's proof seems to run in ZF+DC, so it looks like it is consistent that every $2$-coloring of $[\mathbb{R}]^2$ has a homogeneous set of size $2^{\aleph_0}$.
I just realized that I misread your first question. The answer to that question is surely yes, but I don't have a handy model to show you right now. (I'll try to find one later.)
By the way, if you want to color triples of reals, you can't always get a perfect homogeneous set. For triples, this is illustrated by the coloring $$c(x,y,z) = \begin{cases} 0 left\{\begin{array}{cc}0 & \text{when $mbox{when }2y < x+z$} x+z \cr \ 1 & \text{when $mbox{when }2y \geq x+z$} \end{cases}$$ x+z\end{array}\right.$$ where $x < y < z$. But Galvin showed that you can always get a perfect set whose triples assume at most two colors. Blass later showed that for colorings of $n$-tuples, you can always get a perfect set that takes on at most $(n-1)!$ colors and that this is best possible.
