If you could make it down to the core, then yes, you would probably be able to experience a "solid surface" (where I put that in quotes for reasons that should be apparent in a moment).
The question really gets to, though, what you consider to be a "solid" and a "surface" in a gas giant. The issue at hand is what the Wikipedia article stated - as you go down through the atmosphere, you encounter denser and denser material. Gas under incredibly high pressure will start to behave like a liquid, and be as dense or denser than a liquid such as water. There would be no definite point at which you could say the stuff above you is clearly "air-like" while the stuff below you is clearly "water-like," it's a gradient.
You would also get crushed long before you made it anywhere near the core, just like it's only fairly recently that we've been able to build submersible vessels that can go to the deepest parts of the ocean on Earth.
The answer kind of depends on how old you are. At a very introductory level, say, maybe middle school or younger, it's "okay" to refer to Jupiter as a failed star to get the idea across that a gas giant planet is sort of similar to a star in composition. But around middle school and above (where "middle school" refers to around 6-8 grade, or age ~12-14), I think you can get into enough detail in science class where this is fairly inaccurate.
If you ignore that the solar system is dominated by the Sun and just focus on mass, Jupiter is roughly 80x lighter than the lightest star that undergoes fusion. So it would need to have accumulated 80 times what it already has in order to be a "real star." No Solar System formation model indicates this was remotely possible, which is why I personally don't like to think of it as a "failed star."
Below 80 MJ (where MJ is short for "Jupiter masses"), objects are considered to be brown dwarf stars -- the "real" "failed stars." Brown dwarfs do not have enough mass to fuse hydrogen into helium and produce energy that way, but they do still produce their own heat and glow in the infrared because of that. Their heat is generated by gravitational contraction.
And Jupiter also produces heat through both gravitational contraction and differentiation (heavy elements sinking, light elements rising).
Astronomers are not very good at drawing boundaries these days, mostly because when these terms were created, we didn't know of a continuum of objects. There were gas giant planets, like Jupiter and Saturn, and there were brown dwarf stars, and there were full-fledged stars. The line between brown dwarf and gas giant - to my knowledge - has not been drawn. Personally, and I think I remember reading somewhere, the general consensus is that around 10-20 MJ is the boundary between a gas giant planet and brown dwarf, but I think it's fairly arbitrary, much like what's a planet vs. minor planet, Kuiper belt object (KBO) or asteroid.
So during Solar System formation, was there a chance Jupiter could have been a star and it failed ("failed star!") because the mean Sun gobbled up all the mass? Not really, at least not in our solar system. But for getting the very basic concept across of going from a gas giant planet to a star, calling Jupiter a "failed star" can be a useful analogy.
Best Answer
The reason is electron degeneracy pressure.
The cores of giant planets are dense enough that the electrons in the gas occupy about $h^3$ of phase space each. The Pauli exclusion principle means that they cannot all occupy low energy/momentum states. This means that even at relatively cool temperatures the gas can still exert considerable pressure due to the momenta of the electrons.
A degenerate gas behaves in an anti-intuitive way when it supports a star or planet. A simple argument is the following.
The gravitational potential $\Omega$ and internal pressure $P$ of a planet in equilibrium are related by the virial theorem. $$ \Omega = -3 \int P\ dV,$$ The pressure of a completely degenerate electron gas is proportional to density $\rho$ to the power of 5/3; i.e. $P \propto \rho^{5/3}$ and does not depend on temperature. This is quite a "hard equation of state - the planet becomes difficult to compress.
If we assume the planet has constant density - a terrible approximation, but good enough for a dimensional analysis, then $$ -\frac{3GM^2}{5R} = - 3 \int \frac{P}{\rho}\ dM \propto -3 \rho^{2/3} \int dM,$$ where $M$ is the mass of the star and $\int dm = M$. Substituting $\rho =3M/4\pi R^3$ for the average density, we can easily see that $$ R \propto M^{-1/3}$$ i.e. a more massive star supported by degeneracy pressure is actually smaller, though the dependence on mass is weak.
Now the centres of giant (exo)planets are not completely degenerate, and their outer layers are not really degenerate at all, so this strange behaviour is somewhat moderated. But nevertheless there exists a broad range of planetary masses, from below a Jupiter mass up to tens of Jupiter masses where we expect the radius of the planets to be roughly similar.
The plot below shows some theoretical models compared with some observations from Chabrier et al. (2008). This covers both stars and planets. Notice how the radii of low-mass stars basically decrease (proportional to mass) as the mass decreases and hence $\rho \propto M^{-2}$. But these are supported by perfect gas pressure. As we approach the brown dwarf regime and higher internal densities the electrons become (partially) degenerate and the character of the curves changes and flattens.
Data for transiting exoplanets is also shown. They show a diversity of radii at a given mass that is not completely explained at the current time. Some of it is almost certainly due to irradiation by the parent star (these are almost all "hot Jupiters"). But there may also be composition effects.
EDIT: In response to Steve Everill's points
Note that the $R \propto M^{-1/3}$ behaviour applies roughly between about a few Jupiter masses and 70 Jupiter masses. At lower masses there are various interactions with the ions, Thomas-Fermi corrections etc. that change the ideal degenerate gas behaviour and flatten the relation. This means that when we plot density versus mass for exoplanets, we find that density is proportional to mass (i.e. that the radius is roughly constant). See below - data extracted from exoplanets.org. Below a tenth of a Jupiter mass, the equation of state become much more incompressible and the behaviour changes again.
For normal low-mass stars, the central temperature does not vary a great deal. It is set by the ignition of the pp-chain. Thus the central pressure is $\propto \rho$ for a perfect gas. If you insert this into the treatment I gave above for degenerate stars you find that $R \propto M$ and indeed, the average density of low-mass stars higher.