**This answer is incomplete so far.** I'm pressed for time, so I'll have to improve it later. However, I've worked out a few parameters for you to give you a better idea of the system. I'll add more of the necessary details later.

Let's work out some factors.

 - **Luminosity**

 You gave the radius of the inner edge of the habitable zone as 1.976 AU and the outer edge as 2.808 AU. From this, we can calculate the luminosity of the star. There's an explanation on [Planetary Biology](http://www.planetarybiology.com/calculating_habitable_zone.html). The formulae are
$$r_i=\sqrt{\frac{L_{\text{star}}}{1.1}}$$
$$r_o=\sqrt{\frac{L_{\text{star}}}{0.53}}$$
Plugging in your numbers, I get a luminosity of
$$4.295 L_{\odot}\text{ (inner radius)}$$
$$4.179 L_{\odot}\text{ (outer radius)}$$
I'll average those, giving us a luminosity of $4.237$ times the luminosity of the Sun. But a [P-type orbit](http://en.wikipedia.org/wiki/Circumbinary_planet) is around two stars, as you said, so we divide by two to get an average luminosity of $2.112$ solar luminosities. We can assume that the two stars are similar because they most likely formed together, and have similar properties.

 - **Mass**

 The [mass-luminosity relation](http://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation) can tell us the masses of the stars. It is
$$\left(\frac{L}{L_{\odot}} \right)=\left(\frac{M}{M_{\odot}} \right)^a$$
The stars likely have masses similar to the Sun, so we can assume $a \approx 4$. The left side is $4.179$. We write
$$4.179^{\frac{1}{4}} \times M_{\odot}=M\approx 1.430M_{\odot}$$
So each star is about $1.430$ solar masses, leaving a combined mass of $2.860$ solar masses.

 - **Orbital period of the gas giant**

 [Kepler's Third Law](http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Third_law) tells us that
$$T=\sqrt{\frac{4 \pi ^2}{GM_{\text{star}}}r^3}$$
Here, $M_{\text{Star}}$ is actually the mass of both the stars. If the radius is in the middle of the zone (at about $r=2.392$ AU)
$$T=\sqrt{\frac{4 \pi}{6.673 \times 10^{-11} \times 5.689 \times 10^{30}}(3.578 \times 10^{11})^3}=6.902 \times 10^7 \text{ seconds}= 800 \text{ days}$$

 - **Orbital radius of the moon**

 Here we just go in reverse. We do need the mass of the gas giant, though - so going from the graph on [TimB's answer here](http://worldbuilding.stackexchange.com/questions/4729/is-jupiter-sized-planet-plausible-in-habitable-zone/4733#4733), I'll pick about 5 Jupiter masses, or $9.49 \times 10^{27}$ kilograms. The period will be in between the values you said, so about 52.5 days, which is $4.536 \times 10^6$ seconds. We put this all in and get
$$r=\left( \frac{6.673 \times 10^{-11} \times 9.49 \times 10^{27}}{4 \pi ^2}(4.536 \times 10^6)^2 \right)^{\frac{1}{3}}=6.911 \times 10^{6} \text{ kilometers}$$
Obviously, it's still in the habitable zone. But it's *far* away - although that's because the gas giant is so massive. You may want to opt for a shorter period.

 This system setup appears to be viable if you move the moon closer to the gas giant, giving it a smaller orbital period.

 - **Tidal Locking**

 The formula for [the time it takes for a satellite to be tidally locked](http://en.wikipedia.org/wiki/Tidal_locking#Timescale) is
$$t \approx \frac{wa^5IQ}{3Gm_{\text{planet}}^2k_2R^5}$$
The factors are described on the Wikipedia page. Here, we can say that $I \approx 0.4m_sR^2$, so
$$t \approx \frac{0.4 wQR^2a^6}{3Gm_{\text{planet}}^2k_2r^5}$$
Since
$$k_2 \approx \frac{1.5}{1+\frac{19 \mu}{2 \rho gR}}$$
and $g \approx \frac{Gm_s}{R^2}$,
$$k_2 \approx \frac{1.5}{1+\frac{19 \mu R}{2 \rho Gm_s}}$$
$$k_2 \approx \frac{3 \rho Gm_s}{2 \rho GM_s+19 \mu R}$$
$$t \approx \frac{0.4 wQR^2a^6(2 \rho GM_s+19 \mu R)}{9G^2m_{\text{planet}}^2 \rho R^5}$$
With $Q \approx 100$, $\mu = 3 \times 10^{10}$, $R \approx R_{\text{Earth}}$ and $\rho = \rho_{\text{Mars}}$, you can figure out the tidal locking time. I'm in a rush, so I don't have time to do the calculation, but I may include it later.

> How can I determine how far the moon would need to orbit the gas giant, to not be tidally locked?

Tidal locking will occur at some point in time. You can't get around it.

---

You wrote
> How can I figure out how large the gas giant must be in order to capture this moon, and establish a stable orbit?

I don't have time for this at the moment, but I can say that there are a lot of scenarios in which this could take place. However, you can use the theory of [asteroid capture](http://en.wikipedia.org/wiki/Asteroid_capture) to model it. You can figure out the total energy of the soon-to-be moon by using the formula
$$E=-G\frac{M_{\text{star}} m_{\text{moon}}}{2r}$$
to figure out the energy at a point from the star. Use
$$U_{\text{moon due to gas giant}}=-G\frac{M_{\text{gas giant}}m_{\text{moon}}}{r_{\text{moon} \to \text{gas giant}}}$$
to find its potential energy relative to the gas giant.
Give it an orbiting speed using Kepler's third law (and give the gas giant's speed), the place them a certain distance apart and see what happens. The moon will be captured if $$v<v_{\text{escape velocity}}$$
where
$$v_{\text{escape velocity}}=\sqrt{\frac{2GM_{\text{gas giant}}}{r_{\text{moon} \to \text{gas giant}}}}$$
This modeling will be tough, but you can try it using guess-and-check. Again, I'll expand this section later.