The average distance on a sphere of radius $r$ is $\frac{\pi}{2}r$ (which we can get without any integrations because there's as much area at a distance $\frac{\pi}{2}+\theta$ from a given point as there is at $\frac{\pi}{2}-\theta$); so this is
$\bar{d}=\frac{\pi}{2}\sqrt{\frac{A}{4\pi}}=\frac{\sqrt{\pi}}{4}\sqrt{A}\approx 0.443 \sqrt{A}\;.$
The average distance on the flat manifold $(\mathbb R / a\mathbb Z)^2$ is $\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)a$ (the average Euclidean distance between the point $(a/2,a/2)$ and a random point in $[0,a]^2$), and the area of that manifold is $a^2$, so that makes
$\bar{d}=\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)\sqrt{A}\approx 0.383 \sqrt{A}\;.$
So it's not true that a flat manifold has higher average distance than a sphere; it's the other way around. I suspect the comparison you mean is between a sphere and a flat disk; but a disk is a manifold with boundary, not a manifold, and the higher average distance is due to the boundary, not to the flatness.
[Edit:] Here's the calculation of the average distance in the flat case, as requested. [This is a corrected version.]
We'll need an integral of the form $\sqrt{x^2+c^2}$ several times, so I'll do that in general form first, using the substitution $x=c\sinh u$:
$ \begin{eqnarray} \int\sqrt{x^2+c^2}\mathrm dx &=& \int c\sqrt{1+\sinh^2 u}\;c\cosh u\mathrm du \\ &=& c^2\int\cosh^2u\mathrm du \\ &=& \frac{c^2}{2}(u+\sinh u\cosh u) \\ &=& \frac{c^2}{2}\left(\sinh^{-1}\frac{x}{c}+\frac{x}{c}\sqrt{1+\left(\frac{x}{c}\right)^2}\right) \\ &=& \frac{1}{2}\left(c^2\sinh^{-1}\frac{x}{c}+x\sqrt{x^2+c^2}\right)\;. \end{eqnarray} $
Then the integral over the distance in the primitive cell of a square lattice with $a=2$ is
$ \begin{eqnarray} \int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy &=& 4\int_0^1\int_0^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy \\ &=& 4\int_0^1\left[ \frac{1}{2}\left(y^2\sinh^{-1}\frac{x}{y}+x\sqrt{x^2+y^2}\right) \right]_0^1 \mathrm dy \\ &=& 2\int_0^1 \left(y^2\sinh^{-1}\frac{1}{y}+\sqrt{y^2+1}\right) \mathrm dy\;. \end{eqnarray} $
We can deal with the first term by integrating by parts twice:
$ \begin{eqnarray} \int y^2\sinh^{-1}\frac{1}{y}\mathrm dy &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{1}{\sqrt{1+(1/y)^2}}\mathrm dy\right) \\ &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{y}{\sqrt{y^2+1}}\mathrm dy\right) \\ &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}-\int \sqrt{y^2+1}\mathrm dy\right)\;. \end{eqnarray} $
Putting everything together, we get
$ \begin{eqnarray} \int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy &=& 2\left\{\frac{1}{3}\left[ y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1} \right]_0^1 +\left(1-\frac{1}{3}\right) \int_0^1\sqrt{y^2+1}\mathrm dy\right\} \\ &=& \frac{2}{3}\left[ y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1} +\sinh^{-1}y+y\sqrt{y^2+1}\right]_0^1 \\ &=& \frac{4}{3}\left(\sinh^{-1}1+\sqrt{2}\right)\;. \end{eqnarray} $
This has to be divided by the area $2^2$ to get the average distance for $a=2$, and then by $2$ to get the average distance for $a=1$, since the average distance scales linearly with $a$; that leads to the above result.
For the average distance on the quotient of the plane with respect to a hexagonal lattice, we can use symmetry to restrict the calculation to half of one of the $6$ equilateral triangles. The integrals are rather complicated to work out by hand, but can be done analytically; WolframAlpha gives
$ \int_0^{\sqrt{3}/2} \int_0^{1-y/\sqrt{3}} \sqrt{x^2+y^2}\mathrm dx \mathrm dy =\frac{4+\log 27}{32\sqrt{3}} $
for side length $a=1$. The area of one of these half-triangles is $\frac{\sqrt{3}}{8} a^2$, so the total area of the manifold is $\frac{3\sqrt{3}}{2}a^2$, and the average distance comes out as
$\bar{d}=\left(\frac{4+\log 27}{32\sqrt{3}}/\frac{\sqrt{3}}{8}\right)a=\frac{4+\log 27}{12}\sqrt{\frac{A}{3\sqrt{3}/2}}\approx 0.370 \sqrt{A}\;,$
so yasmar's manifold indeed slightly improves on the one using a square lattice.