Halzen, Martin: Quarks and Leptons
has a good introduction into that matter in Chapter 2.
$|uuu\rangle$ is actually an abbreviation of the idea that we keep an eye on three quarks in particular order, call them "first", "second" and "third"; and then first quark is in $u$ flavour state, of $SU(2)$ or $SU(3)$ flavour group, that is,
$$\left(\begin{array}{c}1\\0\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{c}1\\0\\0\end{array}\right),$$
and second and third quarks are in the same state (so we have to take a tensor product of three such columns, like a 3D matrix).
To check the symmetry of this state, we exchange the states of first and second quarks. (In terms of the 3D matrix, we transpose it on 1st and 2nd axes, in terms of the tensor product we permutate its indices.) Since we had (only a 2D slice is shown)
$$\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0\end{array}\right),$$
our state is immune to such exchange, and we thus call it symmetric. Other possible outcomes would be antisymmetric, or neither of these two symmetries. For example, $|udu\rangle$ is neither symmetric nor antisymmetric, and $\tfrac{1}{\sqrt{2}}(|udu\rangle-|duu\rangle)$ is antisymmetric (with respect to two first quarks only, for simplicity).
If we take into account other variables than just flavour, then we should exchange all variables of two quarks in order to "interchange the quarks".
As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this decay is suppressed.
Best Answer
The (orbital) wave function for $l=1$ doesn't just "have to be" antisymmetric. It demonstrably "is" antisymmetric. The relevant part of this wave function is completely determined, it's a particular function, so we may see whether it's symmetric or antisymmetric and indeed, it's the latter.
Two particles – in this case two pions – orbiting each other are described by a wave function of the relative position $$\vec r = \vec r_1 - \vec r_2 $$ Writing $\vec r$ in spherical coordinates, a well-defined $l,m$ means that the wave function factorizes to $$ \psi (\vec r) = \psi_r(r)\cdot Y_{lm}(\theta,\phi)$$ The angular dependence simply has to be given by $Y_{lm}$ because $Y_{lm}$ is, up to the overall normalization, the only wave function with the right angular momentum and its $z$-component being $l,m$.
But $Y_{lm}$ is easily seen to be an odd function of $\hat r$ for odd $l$ and even function of $\hat r$ for even $l$. In particular, $Y_{00}$ is a constant while $Y_{10}$ and $Y_{1,\pm 1}$ are proportional to $z$ and $x\pm iy $ on the unit sphere, respectively.
These functions proportional to $x,y,z$ are clearly odd functions of $\hat r$, so they change the sign under $\vec r\to -\vec r$ which is the sign flip equivalent to $\vec r_1\leftrightarrow \vec r_2$. This odd parity is also called antisymmetry.