I think you got the sign of the exponential in Schwartz's (28.8) wrong here, which I reverse, but I am cavalier with signs and factors, as I assume you just want to see the principle (the forest, not the trees).
The key point is that the SSB current is basically always of the form $j_\mu(x)\propto F \partial_\mu \pi (x) +...$ where the omitted terms are of higher order in the fields (and so do not contribute to the particle state defined); and, of course, group indices have been omitted as well.
This is codified in the appellation "Nambu-Goldstone nonlinear realization of the symmetry". This is the only way to have the v.e.v. of the transform of this Goldstone field not vanish, while the v.e.v.s of all the fields themselves vanish--after shift redefinitions. (See this answer .) It is illustrated later on in (28.13).
The Fourier transform of the Goldstone field π is
$$
\pi(\vec{\bf{p}}) = \frac{-2i}{F}\int d^3x ~e^{i\,\vec{\bf{p}}\cdot\vec{\bf{x}}}j^0(x),
$$
with
$$
[H,\pi(\vec{\bf{p}})] = E(\vec{\bf{p}}) ~ \pi(\vec{\bf{p}}) .
$$
The SSB charge Q is essentially the space integral of the canonical momentum of π, so that $\langle \Omega |[Q,\int d^3 p ~ \pi(\vec{\bf{p}})]|\Omega\rangle \propto F\neq 0 $, cf. his (28.9).
As a consequence, decompressing your text's one liner, $$H|\pi(\vec{\bf{p}})\rangle= ( [H,\pi(\vec{\bf{p}})] + \pi(\vec{\bf{p}}) H ) ~|\Omega\rangle = ( E(\vec{\bf{p}}) + E_0 )\pi(\vec{\bf{p}})|\Omega\rangle= ( E(\vec{\bf{p}}) + E_0 )|\pi(\vec{\bf{p}})\rangle .$$
Still, for the more conventional general proof of the theorem, bypassing your perceptual snag, consider Kibble's summary.