Which is a stronger acid: H3O+ or HCl?
Here is my opinion, which may be subject to errors and corrections are very welcome: 

Using the definition of Brønsted, acidity is the tendency to give off protons. This, unlike the definition of Arrhenius, is not limited to aqueous solutions. However, if you do have an aqueous solution of an acid, something interesting happens: any acid HAc (or base B) stronger than H$_3$O$^+$ (or OH$^-$) completely dissociates via: 

$$\ce{H2O + HAc &lt;=&gt; H3O+ + Ac-}$$ or 
$$\ce{H2O + B &lt;=&gt; OH- + BH+}$$

So the maximum acidic compound in water is the hydronium ion H$_3$O$^+$; you couldn't, for example, differentiate aqueous hydrochloric and perchloric acid solutions by pH. To rank very strong acids by their acid strength, non-aqueous solutions are used to determine the dissociation constants and the data obtained is then transferred approximately onto water as a solvent. Also note that we are talking about solutions of acids. 

Specifically for HCl, we can denote the pairs of conjugated acids and bases: 

$$\ce{H2O ~(~base~1) + HCl ~(~acid~2) &lt;=&gt; H3O+ ~(~acid~1) + Cl- ~(~base~2)}$$

So we can compare the strengths of the two acids: $\ce{HCl}$ and $\ce{H3O+}$. We know that this equilibrium is shifted to the right, so hydrochloric acid is stronger.

Very late edit to try and answer comment by MukulGoyal (unfortunately, he didn't elaborate on his questions), by number: 


Is the definition of acidity the tendency to give H+ ions, or
H3O+ ions? That seems to matter here because in first case H3O+ will
not even be acidic in water as the equilibrium H2O + H+ ---» H3O+
has a very high equilibrium constant. 
Similarly even if the solvent is NH3, the equilibrium constant of
NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason,
and hence it would actually be a very weak acid. 
Does your answer imply that any acid which has a Ka greater than
1 will be a stronger acid than H3O+?


First of all, please remember that equilibria are dynamic. 

1) As I've mentioned, acidity, according to Brønsted, is the tendency to give off $\ce{H+}$. In $\ce{H2O}$, $\ce{H+}$ are solvated, on average, by 6 water molecules, giving rise to species like $\ce{H3O+(H2O)6}$ (for more information, check out Wikipedia and this). 

1) and 2) you seem to be talking about the dissociation of $\ce{H3O+}$. However, you're using terms like "very high" and "very low", so that, in my opinion, becomes kind of a misleading discussion. 

When you mention 

$$\ce{H2O + H+ &lt;=&gt; H3O+}$$ 

you need to remember that the proton has to come from somewhere. In fact, what you are talking about there is

$$\ce{H2O + H2O &lt;=&gt; H3O+ + OH-}$$ 

which is the autoprotonation of water. Let's just calculate that equilibrium constant you talk about all the time: 

$$K = \cfrac{c(\ce{H3O+})~ \cdot~ c(\ce{OH-}) }{c^2~(\ce{H2O})}$$

Well, guess what? The concentration of water (c = n/V = m / (V * M) = $\rho$ / M = 55,51 mol/L) remains about constant for pure water and diluted solutions (because not many water molecules form ions), so it is brought into the constant, yielding

$$K_w = c(\ce{H3O+})~ \cdot~ c(\ce{OH-})$$

$K_w$ depends on temperature and equals $10^{-14} $ mol$^2$ / L$^{2}$ at 25 °C. It remains constant in pure water and also in diluted solutions. 

In pure water where $c(\ce{H3O+}) = c(\ce{OH-})$, we then get 

$$K_w = c^2(\ce{H3O+})$$ 

and 

$$c(\ce{H3O+}) = 1.0 * 10^{-7} ~mol~ / ~L~ $$

which essentialy quantifies what you, as I suspect, were talking about, in that it tells you the concentration of $\ce{H3O+}$ in the autoprotonation equilibrium for pure water. By the way, because pH is defined like so: 

$$pH = -lg ~c~(\ce{H3O+})$$

at pH = 7 (neutral), $\ce{H3O+}$ and $\ce{OH-}$ are equal. In diluted solutions, when they are not equal, you can calculate one using the other. 

Also, remember that in the Brønsted–Lowry theory acids and bases are defined by the way they react with each other, as mentioned in Wikipedia. So far, we have only looked at water and its autoprotonation, no solvents and other molecules, which have their own acidies and basicities. 

2) As for the reaction you mentioned in ammonia as a solvent, well, first of all, $\ce{H3O+}$ needs to come from somewhere (let's imagine we drop some aqueous solution of hydrochloric acid in there). Second, I am not sure that constant would actually be as you describe it - I think that here, ammonia (which behaves ampholytic) would act as a base and pretty happily accept a proton. If anyone can provide insight into this, that would be great. 

3) Acid strength is defined using $pK_a$, which is given by $pK_a$ = -lg $K_a$. From my experience, if $pK_a$ is defined via concentration, then $pK_a$ of hydronium ions is 0, and if it is defined via activity, it is -1.7 (-1.74, to be more exact; similarly, for $\ce{OH-}$, the analogous $pK_b$ is 14 via concentration and 15.7 via activity). 

Coming back to your question about "Ka greater than 1", that seems about right, because -lg (1) = 0. 

I refer you to an acid strength table, for instance, here. Also, I'd like to refer you to this excellent discussion on the matter, where, by the way, orthocresol lent a hand. 

Also, just to add this, strong acids have $pK_a$ &lt; 0, while weak acids have $pK_a$ > 0 (again, this is via concentration - for example, here in Wikipedia they use -1.74). 

Everyone, corrections are very welcome!