Which overlap is stronger, and why?
An atom's wavefunctions (orbitals) are orthogonal. This follows from the fact that the Hamiltonian is Hermitian, i.e eigenfunctions (wavefunctions) of quantum mechanical operators are orthogonal.  If they are on different atoms then they are not orthogonal and the overlap clearly depends on how far apart the atoms are.

Mathematically orthogonality means that (in one dimension) two wavefunctions $ \psi_i$ and $\psi_j$ satisfy the condition 
$$\int_{-\infty}^{\infty} \psi_i\psi_jdx =\delta_{ij}$$

where $\delta_{ij} = 1 \text{ if } i=j \text { else } 0$ and is the Kroncker delta. (In this case if $i=j$ then the wavefunctions are also normalised as the product is $1$, this joint condition, orthogonal &amp; normalised  is sometimes called 'orthonormal')

In the case of atomic orbitals there are three quantum number, the principle one n and angular ones l and m. The orthogonality condition is then
$$ \int_0^{\infty}r^2\int_0^{\pi}\sin(\theta) \int_0^{2\pi}\psi^*_{n'l'm'}(r,\theta,\phi)\psi_{nlm}(r,\theta,\phi)~dr ~d\theta ~d\phi=\delta_{nn'}\delta_{ll'}\delta_{mm'}$$

In your examples $n=2$ and $n'=3$ so the orbitals are orthogonal on these grounds alone.

As a check the 2s orbital has the form 
$$ \psi_{2s} = A(2-\sigma)\exp(-\sigma/2) $$
where A contains only constants and $\sigma=Zr/a_0$ where Z is the atomic number, $a_0$ the Bohr radius and r the radial distance. One 3p orbital ($l=1, m=0$) has the form
$$\psi_{3p} = A'\sigma(6-\sigma)\exp(-\sigma/3)\cos(\theta)$$

Integrating the product over the angles $\theta$ which is  the integral $\int_0^{\pi}\cos(\theta)d\theta $. We can see that this integral is zero from a plot of cosine.

Edits: 

As the question is now amended, the answer above holds for a single atom, but when there are two atoms then the overlap depends on their separation as the wavefunctions are no longer orthogonal because they have different coordinate origins, one on one atom the other on the other. 

The overlap integral is written as
$$S= \int_{-\infty}^{\infty} \psi_i\psi_j~d\tau$$

where $\tau$ is the set of coordinates. Even in the simplest case of $\ce{H2^+}$ the coordinates needed are prolate spheroidal ones so to give an example it is simplest to use a one dimensional example since the principle is the same but technically more complex in 3D.

Starting with H atom 1s orbitals 
$$ \psi_{1s} = a_0^{-1/2}\exp(-x/a_0)$$

if the atoms a and b are separated by 2R then the wavefunctions are 

$$ \psi_a = a_0^{-1/2}\exp(-|x-R|/a_0) \\
\psi_b = a_0^{-1/2}\exp(-|x+R|/a_0)$$

and the overlap is the integral 
$$ S=\int_{-\infty}^{\infty} \psi_a\psi_b dx =a_o^{-1}\int_{-\infty}^{\infty} \exp(-|x-R|/a_0)\exp(-|x+R|/a_0) dx$$

which evaluates to $$S=\frac{(a_0+2R)\exp(-2R/a_0)}{\pi a_0^3}$$

The wavefunctions with $a_0=1$, (and $R=1$ for the first two plots), $\psi_a$ and $ \psi_b$ vs. x  overlap $\psi_a.\psi_b$ vs. x and overlap integral S vs. the atom separation R are shown in the figures.