It seems that Britannica is more accurate.
The Britannica values of $g$ can be traced back to Table 2 of this paper: The atmosphere of Saturn - an analysis of the Voyager radio occultation measurements, Lindal et al (1985). As stated in the title, these values are derived from the Voyager measurements.
At first, I thought that the wiki/Nasa value was based on the more recent Casini measurements. However, it turns out that this value actually comes from a simplified (and inaccurate) calculation.
I haven't found an explicit recent value of $g$ in the literature, but I did find out how to calculate it: on page 3 of the paper Interior Models of Saturn: Including the Uncertainties in Shape and Rotation (Helled & Guillot, 2013), it is stated that the effective potential of a rotating planet is
$$
U(r,\varphi) = \frac{GM}{r}\!\left(1 - \sum_{n=1}^\infty\left(\frac{r_\text{s}}{r}\right)^{2n}J_{2n}P_{2n}(\sin\varphi)\right) + \frac{1}{2}\omega^2r^2\cos^2\varphi,
$$
where $M$ is the mass of the planet, $r_\text{s}$ is a reference equatorial radius, $\varphi$ is the latitude, $\omega=2\pi/P$ is the angular velocity (and $P$ is the rotation period), $P_{2n}(x)$ are Legendre polynomials and $J_{2n}$ are coefficients that express the deviation of the planet from a perfect sphere. The gravitational acceleration $\mathbf{g}$ is then a vector with components
$$
\begin{align}
g_r(r,\varphi) &= -\frac{\partial U}{\partial r},\\
g_\varphi(r,\varphi) &= -\frac{\partial U}{\partial \varphi}.
\end{align}
$$
Since $g_\varphi=0$ at the poles and the equator, let's focus on $g_r$ only. We find
$$
g_r = \frac{GM}{r^2}\!\left(1 - \sum_{n=1}^\infty(2n+1)\left(\frac{r_\text{s}}{r}\right)^{2n}J_{2n}P_{2n}(\sin\varphi)\right) - \omega^2r\cos^2\varphi,
$$
which is the same as the formula given in the Appendix of Lindal et al (1985). Now let's plug in some numbers: from Helled & Guillot we have
$$
\begin{align}
r_\text{s} &= 60,330\;\text{km},\\
J_2 &= 16,290.71\times 10^{-6},
\end{align}
$$
ignoring the smaller higher-order coefficients $J_4$, $J_6$,...
In Table 2 of Anderson & Schubert (2006) we find:
$$
\begin{align}
a &= 60,357.3\;\text{km},\\
q &= 0.158904,\\
P &= 10^\text{h}32^\text{m}35^\text{s} = 37,955\;\text{s},
\end{align}
$$
where $a$ is the equatorial radius and $q$ the so-called smallness parameter. From this, we obtain
$$
\begin{align}
\omega &= 2\pi/P = 1.65543\times 10^{-4}\;\text{rad}\,\text{s}^{-1},\\
GM &= \omega^2a^3/q = 3.79207\times 10^{16}\;\text{m}^3\,\text{s}^{-2}.
\end{align}
$$
Finally, we need the equatorial and polar radius at 1 bar. According to the IAU report on cartographic coordinates and rotational elements (2006), pag. 173, we find
$$
\begin{align}
r_\text{eq} &= 60,268\;\text{km},\\
r_\text{pol} &= 54,364\;\text{km}.
\end{align}
$$
Thus, at the equator we have (ignoring higher-order terms)
$$
\begin{align}
g_r(r_\text{eq},0^\circ) &= g_1 + g_2 + g_3\\
&= \frac{GM}{r_\text{eq}^2} + \frac{GM}{r_\text{eq}^2}\frac{3}{2}\!\frac{r_\text{s}^2}{r_\text{eq}^2}\!J_2 - \omega^2r_\text{eq},
\end{align}
$$
so, the gravitational acceleration comes from three contributions: the gravity of a spherical planet, a correction due to non-sphericity and a centrifugal influence due to the rotation. We obtain
$$
\begin{align}
g_1 &= 10.44\;\text{m}\,\text{s}^{-2},\\
g_2 &= 0.256\;\text{m}\,\text{s}^{-2},\\
g_3 &= -1.652\;\text{m}\,\text{s}^{-2},
\end{align}
$$
so that the acceleration in a frame rotating with the planet is
$$
g_r(r_\text{eq},0^\circ) = 9.04\;\text{m}\,\text{s}^{-2}.
$$
Contributions of the $J_4$, $J_6$ terms may change the last digit.
Now we see where the 10.44 value comes from: it is only the spherical term, incorrectly ignoring the effects of non-sphericity and rotation. At the poles, we have
$$
\begin{align}
g_r(r_\text{pol},90^\circ) &= g_1 + g_2\\
&= \frac{GM}{r_\text{pol}^2} - \frac{GM}{r_\text{pol}^2}\!3\!\frac{r_\text{s}^2}{r_\text{pol}^2}\!J_2,
\end{align}
$$
which results in
$$
\begin{align}
g_1 &= 12.83\;\text{m}\,\text{s}^{-2},\\
g_2 &= -0.772\;\text{m}\,\text{s}^{-2},\\
g_r(r_\text{pol},90^\circ) &= 12.06\;\text{m}\,\text{s}^{-2}.
\end{align}
$$
So the equatorial acceleration is indeed about 75% of the acceleration at the poles. Conclusion: Britannica is more accurate, although the values are not up-to-date.
