The neutron is composed of quarks (and gadzillions of antiquarks) both left and right-chiral: the quark mass terms connect them to each other and both species have a role to play.
In addition, you readily see in the PDG review that the weak hypercharge connects to both species in a magnificently lopsided way (Feynman used to call it "cockeyed"). So, by necessity, the neutron cannot be weak neutral.
For instance, the quark effective Lagrangian in the SM has a term like
$$
W^+_\mu J^{\mu ~+}= W^+_\mu \bar u \gamma^\mu P_L d ,
$$
and, as you indicated, this elicits the decay of the free neutron, since
$$
\langle p| J^+ |n\rangle \neq 0 ,
$$
as you may read up on in standard texts, like the one by M. Schwartz.
But the range of the weak interaction represented by this is 0.1% the size of the neutron, one fermi, so it all happens deep inside it, if you wanted a dream metaphor. It's hard to see what an "external weak force" would be like. (It might be a virtual pseudoscalar coupling, like $K^-$, but pay no mind...) So the weak isospin charge of the neutron is a marshmallow mess computable in current algebra, and it is non vanishing. (Cf. Erler & Su Progress in Particle and Nuclear Physics
Volume 71, July 2013, Pages 119-149.)
With similar arguments, you may see the neutral current coupling of the neutron is nonzero, and it probes both L and R quarks inside it, because the weak hypercharge U(1) couples nontrivially to both. But the clever Weinberg combination of electromagnetism remains unbroken, and a Ward identity ensures that the very long distance interactions of a photon with the neutron vanish: zero charge. (On shorter distances, there are magnetic interactions, as G.Smith's answer points out.)
- Comment on extra comment
Suppose we do not consider the quark sea. Just the representation theory of a fermion bound state. Does the right-handed neutron as a bound state count as gauge-neutral to all gauge forces?