What is $\pi_2(\mathbb{R}^2 - \mathbb{Q}^2)$?

I'm working through Hatcher book and done $\pi_1(\mathbb{R}^2 - \mathbb{Q}^2)$ is uncountable. It's easy to see that it's true as you can imagine only non trivial maps contract in the space.

But, was wondering has anyone worked out $\pi_2(\mathbb{R}^2 - \mathbb{Q}^2)$. $\pi_2(\mathbb{R}^3 - \mathbb{Q}^3)$ I imagine this isn't that hard, shouldn't it be uncountable aswell.

Just doing a project on higher homotopy theory, but fundamental group stuff seems hard already. Like $\pi_2$ is that just a sphere doing a weird thing and looping back on itself. I know the definition but can't really see it. Plus the calculation aren't that easy.

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$\pi_2(\mathbb R^3\setminus\mathbb Q^3)$ is certainly uncountable. $\pi_2(\mathbb R^2\setminus\mathbb Q^2)$ is most likely trivial, but I don't see a quick argument. – Grumpy Parsnip Feb 16 '12 at 18:42
I find the standard definition of $\pi_2$ fairly unenlightening. The definition that to me works much better is that it is the group of homotopies between null-homotopic paths from a basepoint to itself, up to homotopy. In other words, it's the fundamental group of the loop space (en.wikipedia.org/wiki/Loop_space). – Qiaochu Yuan Feb 16 '12 at 18:52
These things can be counter-intuitive. The hawaiian earring space is a retract of $\mathbb R^2\setminus\mathbb Q^2$, and if you look at the 2D version of the hawaiian earring space, $X$, Barratt and Milnor showed that $H_3(X)\neq 0$. The plane is usually exempt from this sort of exotic behavior, so I'll bet $\pi_2=0$ in that case, but I also think the proof may be quite difficult. – Grumpy Parsnip Feb 16 '12 at 19:26
@QiaochuYuan I will have to look at that in more details. I can see $\pi_1$ and $\pi_2$ seems like a natural generalization. – simplicity Feb 16 '12 at 19:38
@Qiaochu, I find it almost incredible that that definition (which is of course very very nice technically) can be considered more enlightening that the definition of $\pi_2$ as homotopy classes of maps from a sphere! – Mariano Suárez-Alvarez Feb 17 '12 at 2:51

which shows that every subset of the plane has trivial $\pi_k, k\geq 2$. So in particular $\pi_2(\mathbb R^2\setminus \mathbb Q^2)=0$.