Polar molecules have an electric field due to there being a net dipole moment. The electric field will be stronger with greater polarity. The stronger electric field in the polar molecule will interfere with IR radiation stronger than a weaker electric field. But how does the interference between the molecule's electric field and the IR radiation have to do with intensity of IR radiation absorbtion of the molecule? Is there a causal relationship between the two or is there simply a correlation?
To answer your question 'yes' there is a relationship. The process is not quite as you put it however. If a bond has atoms of different electronegativity then obviously there is a dipole formed and as the molecule vibrates about its mid point the dipole also oscillates since the nuclei move. The electric field of the radiation can interact with the molecules's oscillating dipole and only if the frequency is very, very close to that of the vibration energy can be transferred from the radiation to the molecule. This 'resonance' results in an absorption band at the vibrational frequency, and a molecule with one extra quanta of vibrational energy.
This is a phenomenological description, in a quantum description the probability of absorbing energy and moving from one vibrational level (stationary state) to another is proportional to the square of the strength of the radiation's electric field $\epsilon$ and the square of the transition moment $M$ of the vibration. Thus probability of absorption $\sim \epsilon^2M^2$. The electric field squared is proportional to the light intensity; no light no absorption, obviously.
The transition moment is proportional to the dipole $\mu$ as $M = \int \psi_i^*\mu \psi_f dx= q\int x\psi_i^* \psi_f dx$ where $\psi$ are the vibrational wavefunctions for the initial and final states in the transition, say v = $0,1$ and $q$ the charge on the electron. This shows that if the dipole is zero then there is no transition, just as in the case of homonuclear diatomic molecules. Additionally this shows that even if $\mu$ is large if the wavefunctions do not overlap (their product is zero) then the integral can be zero. This leads to the selection rules for transitions, which means that not all potential transitions occur.