Are there any (simple) molecules with very different absorption and emission dipole directions?
In part of the naphthalene molecule you draw you indicate a transition moment. This is the direction for absorption into the lowest excited state $S_1$. As naphthalene is highly symmetric the fluorescence is also emitted from a transition moment aligned in the same direction. The second excited state $S_2$ has a transition dipole set perpendicularly to the long axis and so to the $S_1$ direction.  Again excitation and emission would be from similarly aligned transition dipoles if emission was from $S_2$. However, in solution $S_2$ decays very rapidly ( a few picoseconds) to $S_1$ thus the observed absorption and emission dipoles would be at right angles to one another.

The wavefunction (molecule orbital) in an excited state is different to that in the ground state since an electron is excited to an anti-bonding orbital. This means that bond lengths and angles can change a little. If the molecule has polarisable substituents that are asymmetrical placed then the excited state dipole may differ in direction from that of the ground state. However, in molecules such as rose bengal, rhodamine 6G and many dicarbocyanines, which are not strictly symmetric ($C_1$ point group), the absorption and emission dipoles are parallel as determined by experiment.  (Presumably this is because substituents are not sufficiently close in energy to affect the electronic state in a significant way)

How can we know what the angle between absorption and emission is? The probability of absorption is proportional to the cosine squared of the angle $\theta$ between the electric field direction of the light and the transition moment direction, $\approx (\bar e \cdot \bar\mu)^2= e^2\mu^2\cos^2(\theta)$. So we can describe the molecules are being photo-selected, the probability of absorption being greatest when $\theta=0$, i.e. absorption and excitation directions are parallel.   

If we use polarised light to excite the molecules, and assume that it is polarised along the $z$ direction, then the  emission intensity observed through a polariser parallel to the $z$ (or vertical) direction is $I_v\approx (\bar e_z \cdot \bar\mu_z)^2$  and observed at right angles to this ($x$ or horizontal direction)  $I_h\approx (\bar e_z \cdot \bar\mu_x)^2$ where $\bar e_z$ is a unit vector aligned along the electric field and  $\bar \mu_z$ the unit vector component of the fluorescence dipole along the $z$ and $\bar \mu_x$ along the  $x$ direction. The amount of polarisation of the fluorescence is usually measured via the  anisotropy which is defined as 

$$r=\frac{I_v-I_h}{I_v+2I_h}$$

where the denominator is the total emission intensity. 

In fluid solution molecules undergo rotational diffusion and if the molecule is regarded as a sphere, in fact not such a bad approximation in many cases then, the anisotropy becomes $\displaystyle r(t)=r_0e^{-6Dt}$ where $D$ is the rotational diffusion constant. The anisotropy can be measured by recording the fluorescence decay such as by time-correlated single photon counting. Taking the decay back to time zero produces $r_0$, this is the anisotropy that would be observed if the molecules were in rigid solution and effectively measures the angle between absorption and emission dipoles.  

Now this relates directly to your question. The value of $r_0$ can be also be calculated as well as being measured and, again, considering the angle between absorption and emission dipoles $r_0=(3\cos^2(\alpha)-1)/5$ where $\alpha$ is the angle between absorption and emission dipoles. When the angle is zero then $r_0=0.4$ which is easily measured and when at right angles $r_0=-0.2$ so this limits the range of anisotropy. 

(references. Text. M. Daune. Molecular Biophysics publ. OUP 1999. Some original literature; T Tao, Biopolymers v8,p609, 1969, Y. Yguerabide Meth. Enzymology, v26, p498, 172, Chuang &amp; Eisenthal, J. Chem. Phys. v57, p5094, 1972)