This is a very good question.
First let me clarify a point. So far long range entanglement is only defined for gapped quantum systems.
The gapless systems seems always "long range entangled". So the notion is useless.
Do RVB states have long range entanglements? I think the string idea that you mentioned is a very good idea: A string liquid leads to long range entanglements.
Are RVB states string liquid states? Actually, the answer is yes.
We may take a VB configuration as a reference, than the difference between any other VB configuration and the reference VB configuration can be described by a closed string! (See Sutherland, Phys. Rev. B 37 3786 1988; Kohmoto, Phys. Rev. B 37 3812, 1988).
So a RVB state is actually a string liquid! If the dimmers only connect
between A sub-lattice and B sub-lattice, then the strings are orientable
and the corresponding string liquid gives rise to an emergent U(1)
gauge theory. Otherwise, the strings are not orientable
and the corresponding string liquid gives rise to an emergent $Z_2$
gauge theory.
In fact, the situation is a more complicated than the above discussion.
The string liquid from RVB is not an equal weight superposition of all loop
configurations. Different string configurations may have different weights.
So the string liquid from RVB may not be a liquid. It could be a string solid
+ a little fluctuations. In this case, there is no emergent gauge theory
and the corresponding RVB state is not long range entangled.
If a RVB state does correspond to a loop liquid, then the corresponding RVB state is long range entangled.
Garret's presentation of what he calls the "Einstein-Podolsky-Rosen-Garret" paradox, in the 25:00 - 29:00 range of the video you link to, is not sound.
Garret proposes a source of entangled particles which produces the state
$$
\newcommand{\up}{|\!\uparrow⟩}\newcommand{\down}{|\!\downarrow⟩}
|\Psi⟩=\frac{\up\down+\down\up}{\sqrt{2}}
$$
and then sends the particles to distant locations, to be measured at spatially-separated events. His protocol then asks you to 'measure on the left, and look for interference on the right', which can be succinctly phrased as measuring on the $\{\up,\down\}$ basis on the first mode, and on the
$$
\newcommand{\plus}{|+⟩}\newcommand{\minus}{|-⟩}
\left\{\plus=\frac{\up+\down}{\sqrt{2}},\minus=\frac{\up-\down}{\sqrt{2}}\right\}
$$
basis on the second mode. The measurement probabilities in this case are easily seen to be
$$
\newcommand{\bup}{⟨\uparrow\!|}\newcommand{\bdown}{⟨\downarrow\!|}
\newcommand{\bplus}{⟨+|}\newcommand{\bminus}{⟨-|}\newcommand{\bpm}{⟨\pm|}
\left|\bup\bpm|\Psi⟩\right|^2\frac12,
$$
and analogously for $\bdown$, so if you measure on the left there is no interference on the right.
Garret then claims that this can be used for superluminal communication, and this is where he is incorrect.
So far, there's nothing that Alice, who is in control of the first mode, can do to alter the outcome at all - she definitely cannot control which of the two outputs ($\up$ or $\down$) she will get. The only choice she has is whether to measure her system prior to the interference step, or to let the two arms interfere and then measure. What Garret apparently doesn't realize is that even if Alice does let her system produce interference, the other system will not produce interference either. Alice has no way to make Bob's side of the system display interference without sending him classical information at subluminal speeds.
Let me sketch that calculation as it is important to the argument. Suppose both Alice and Bob measure on the $|\pm⟩$ basis, with Alice obtaining $|a⟩=|\pm⟩$ and Bob obtaining $|b⟩=|\pm⟩$. The probability for this outcome is then
\begin{align}
|⟨a|⟨b|\Psi⟩|^2
&=\frac18\left|(\bup+a\bdown)(\bup+b\bdown)(\up\down+\down\up\right|^2
\\&= \frac18\left|a+b\right|^2.
\end{align}
Thus if Alice gets $\plus$ it is certain that Bob will get $\plus$, and ditto for $\minus$, so it seems that Bob does observe interference. However, just because Alice decides that she wants to run her system through the 'recombine' step that doesn't mean that she gets to control which outcome she gets. She will get $\plus$ as often as she does $\down$, which means that so will Bob, and what that looks like to Bob is simply no interference.
Alice's actions, then, have no effect on what Bob observes, and therefore this channel cannot be used for superluminal communication.
This does sort of have a bearing on the Copenhagen Interpretation as described by Garret,
most people think of the Copenhagen interpretation as the idea that measurement causes a physical phenomenon known as “collapse of the wave function” which is non-linear and irreversible, i.e. it causes a quantum superposition to change into a probabilistic mixture of classical states, and that there is an physical difference between these two states.
but it doesn't rule CI out. In particular, this understanding of the Copenhagen Interpretation does force Bob's system to change instantaneously as soon as Alice measures her side of their entangled pair, but Bob's system somehow contrives to (partially) "hide" this inner change of state from any possible measurement, in a way which exactly prohibits superluminal communication.
This bothers some people (it somehow imagines Nature as having an extra set of supernatural powers which it purposefully denies us) but it is not inconsistent with the laws of physics. In particular, this "EPRG" argument does not rule out the Copenhagen Interpretation as "scientifically untenable", as Garret appears to claim.
I'm afraid I won't have time to critique whatever it is he says in the second half of the video, though. Quantum interpretations are a tricky business, and if you make technical mistakes on the underlying mechanics then there's no telling how much of a basis the resulting arguments will have.
Best Answer
I may be able to offer some hints. I am also looking forward to more answers (Ryan please!) because this is definitely a complicated subject, with contributions from people coming from very different backgrounds (hep-th, cond-mat, pure math, etc.).
"Condense" may mean different things to different people. I believe the oldest precise definition was that an operator condenses if it picks up a vacuum expectation value. The physical interpretation is that the operator acquires a macroscopic value, i.e., it becomes something like a constant background, a spacetime filling object.
Note that vevs typically imply that a symmetry is spontaneously broken (and not the other way around!). Indeed, if the operator is charged under a given symmetry, then its condensate breaks the symmetry. Of course there are situations where an operator that is not charged under any symmetry condenses, and in those cases the vev does not break any symmetries, and thus there is no spontaneous symmetry breaking. The simplest example is, I think, the Schwinger model, where the basic meson condenses but there are no symmetries to begin with, so nothing is broken. Of course, Schwinger is gapped and has a unique, trivial vacuum state.
In QCD, the object that typically condenses is the quark bilinear $\psi_L\psi_R$. If this operator has a non-zero expectation value, it breaks the chiral symmetry $SU(N)_L\times SU(N)_R$ down to the diagonal subgroup. This breaking means that there are Goldstone bosons, the pions. There are other objects that could condense beyond the quark meson (e.g., with enough supersymmetry it is known that certain monopoles condense; I believe they break the magnetic $U(1)$ one-form symmetry and hence the Goldstone photon becomes massive and is gapped out, hence the theory becomes confining. But don't quote me on that).
More generally, condensing means that you allow the object to have a macroscopic value. In cond-mat, I believe, this is often achieved, roughly speaking, by dropping certain terms from the Hamiltonian, i.e., those terms that energetically penalize the excitation of the object you want to condense. If the object to condense does not have a term in the Hamiltonian, then it will cost no energy to produce it, and the vacuum state will have a non-zero excitation number for it, i.e., the object has acquired a macroscopic value.
A different notion of condensing is that of gauging. If you gauge a symmetry generated by that operator, then effectively you allow that operator to proliferate, because gauging mean that you identify any two objects that differ by the operator you gauged. Thus, the gauged operator is identified with the vacuum. This is reminiscent of the previous paragraph, since the Hamiltonian must be gauge invariant, i.e., it cannot depend on the object you gauged. (Note that in cond-mat, gauging is often enforced energetically, i.e., you impose the Gauss law by penalizing configurations that violate it).
This last notion is the one used for strings. Condensing a string means gauging the one-form symmetry it generates. Gauging means summing over bundles, i.e., you introduce a mesh of strings, which reproduces the fact that the condensed object acquires a macroscopic value. When you gauge a string, you sum over all possible insertions, and therefore the string appears everywhere: the vacuum consists of a spacetime filling mesh of strings.