It's actually the other way around. That axial rotation of the pions ensures that, given its non-vanishing v.e.v., given by the condensate (assumed to be produced by QCD: a fact!), they therefore must be the Goldstone modes of the SSB of the axial charges. But, first, the chiral condensate is required so as to unleash all this.
Take L=2, for simplicity, and let's be schematic (~) about normalizations, which you may adjust to your satisfaction, in comportance with the conventions of your text.
Let us consider the relevant fermion bilinears and their representation of the $SU(2)_L\times SU(2)_R$. (By the way, the 3 axial $\vec{Q}_A$ do not close to an SU(2), as you wrote, as their commutators close to $SU(2)_V$ instead. Don't ever write this again... Also, $U(1)_A$ is broken explicitly by the anomaly, not spontaneously.) So, the 4 bilinears,
$\bar{\psi} {\psi} $, $\bar{\psi}\gamma_{5} \vec{\tau} {\psi} $ form a quartet of this chiral group formally analogous of the $\sigma, \vec{\pi}$ quartet of the σ-model of the 60s; in fact, they are the QCD interpolating fields for this quartet.
(If you are insouciant about i s and the like, you may think of this scalar-pseudoscalars quartet as a 4-vector $t,\vec{x}$ of the Lorentz group,
as a familiar mnemonic of the combinatorics/groupery to follow: The vector isorotations are analogous to the 3 rotations, and the 3 axials are analogous to the 3 boosts, acting on 4-vectors.)
You then see that
$$
\left[\vec{Q}_{V},\bar{\psi}\psi\right]=0,
$$
$$
\left[\vec{Q}_{A},\bar{\psi}\psi\right]\sim \bar{\psi} \gamma_{5} \vec{\tau} \psi ~,
$$
$$
\left[Q_{a}^{V} ,\bar{\psi} \gamma_{5}\tau_{b} \psi \right] \sim \epsilon _{abc}
\bar{\psi}\gamma_{5}\tau^c\psi ~,
$$
and, crucially, the relation of interest, where you note the $\gamma_{0}\gamma_{5}$ makes all the difference in the combinatorics, as in the σ -model link, above,
$$
\left[Q_{a}^{A} ,\bar{\psi} \gamma_{5}\tau_{b} \psi \right] \sim
\delta_{ab} \bar{\psi}\psi ~.
$$
So the σ is an isosinglet; the axials transform the σ by $\vec\pi$s; the vector transform of the $\vec \pi$s is an isorotation thereof; and the suitable "diagonal" axial transforms of the $\vec\pi$s takes them to the
σ, the QCD analog of the Higgs of the EW interactions, the guy with the v.e.v.
Now take the v.e.v. $
\langle \Omega \lvert ... ... \rvert\Omega \rangle$ of each of the above.
The r.h.sides of the first 3 must vanish; the v.e.v.s of the Goldstone modes are null!
The v.e.v. of the last one does not, but $=v\approx (250MeV)^3$, your relation of interest. QCD just achieves that, by dint of strong dynamics. So, over and above making it impossible for the axials to annihilate the vacuum, it identifies the 3 pions as the Goldstone modes of the 3 SSBroken $\vec{Q}_A$.
In point of fact, you actually see that $\vec{Q}_{A} \rvert\Omega \rangle \sim |\vec{\pi}\rangle$, that is the axial charges pump pions (chiral goldstons) out of the vacuum--- the precursor to PCAC.
A final loose end, lest you might still object that the third relation with vanishing r.h.side would then be moot, $\langle \vec{\pi} \lvert \bar{\psi} \psi \rvert\Omega \rangle=0$; but, no, the pion is orthogonal to the σ, just as it is to the original vacuum "chosen", $\langle \vec{\pi} \vert\Omega \rangle=0$.
(Could insert $\vert\Omega\rangle \langle\Omega\vert$ above and factor out the order parameter, $\langle \vec{\pi} \vert\Omega \rangle v=0$ )
Further related (formally identical) questions might be 281696, and, needless to say, Gell-Mann and Lévy's timeless 1960 σ -model classic.
In a world without EW SSB, pions would, indeed, be perfect massless! Pion masses reflect two different SSBs.
There are two SSB scales involved in the SM: the electroweak spontaneous breaking of $SU(2)_L \times U(1)_Y$, with an order parameter of about 1/4 TeV; and the chiral spontaneous breaking of the three axial generators (not closing into an $SU(2)$, of course) with the quantum numbers of the three pions, with order parameter about 1/4–1/5 GeV, close to $\Lambda_{QCD}$; two completely different phenomena, understood very differently.
There is no sense in talking about pions above the higher scale; but below it, you do have effective quark masses, generated from the Yukawa couplings: the heart of the SM, regardless of how poorly they teach it to you.
So, below that upper scale, and around the lower one's χSSB that P&S detail, there are quark masses, for all practical purposes, and all they say makes eminent sense! Check that all statements are consistent and reasonable in all three regions demarcated by these two scales.
This includes your puzzlement about (pseudo)Goldstone bosons: the Goldstone property refers to the non-linear transformation law of the particles under the SSBroken symmetries. Pions, composite or not, are interpolated by operators transforming in the (Wigner-Weyl, or else) Goldstone mode: and our real world pions choose the latter.
This is why, as you are taught, it is very impractical to think of pions as ordinary hadron bound states, which they also are. Primarily, they are goldstons. In fact, there is a small energy region in-between $\Lambda_{QCD}$/confinement-scale and the above χSSB, where you may consider Goldstone degrees of freedom and constituent massive quarks, the chiral bag models, but angels fear to tread there... You really need not get interested in it.
Best Answer
As stated in your question, the massless QCD Lagrangian $$ \sum_{f= u,d,s} \bar{q}_{Lf} i \gamma^{\mu} D_{\mu} q_{Lf} + \bar{q}_{Rf} i \gamma^{\mu} D_{\mu} q_{Rf}$$ is invariant under $ U(1)_R \times U(1)_L$. That just means that you can multiply the left-handed and right-handed quarks by independent (global) phases: $$U(1)_R \times U(1)_L: q_{Rf} \to e^{i\epsilon_R} q_{Rf}, \quad q_{Lf} \to e^{i\epsilon_L} q_{Lf},$$ where the $\epsilon_L$'s and $\epsilon_R$'s are the same for all f's. The corresponding Noether currents are $$ L^{\mu} = \sum_{f= u,d,s} \bar{q}_{Lf} \gamma^{\mu} q_{Lf}, \quad R^{\mu} = \sum_{f= u,d,s} \bar{q}_{Rf} \gamma^{\mu} q_{Rf}. $$ The Noether currents of $ U(1)_V \times U(1)_A$ are now just linear combinations of those: $$ V^{\mu} = L^{\mu} + R^{\mu} = \sum_{f= u,d,s} \bar{q}_{f} \gamma^{\mu} q_{f}, $$ $$ A^{\mu} = R^{\mu} - L^{\mu} = \sum_{f= u,d,s} \bar{q}_{f} \gamma^{\mu} \gamma_{5} q_{f}. $$ You can check that these two transform as a vector and axial vector respectively. It is useful to remember how Dirac spinors and $\gamma_5$ look in the chiral basis $$ q = \begin{pmatrix} q_L \\ q_R \end{pmatrix}, \quad \gamma_5 = \begin{pmatrix} -I_2 & 0 \\ 0 & I_2 \end{pmatrix} .$$ As we did for the currents, we can also take linear combinations of the generators $$ \alpha=\frac{\epsilon_L + \epsilon_R}{2}, \quad \beta=\frac{\epsilon_R - \epsilon_L}{2}. $$ The $ U(1)_V \times U(1)_A$ acts on Dirac spinors now as follows: $$ U(1)_V \times U(1)_A: q_f \to e^{i \alpha} q_f, \quad q_f \to e^{i \beta \gamma_5} q_f, $$ which is equivalent to the transformation above.