If $F$ is finite, $p$ and $q$ are irreducible, and have the same dagree $d$, then $F[x]/I$ and $F[x]/J$ are isomorphic $F$-algebras, since there is (up to isomorphism) only one extension field of a finite field of any given degree (see for example Birkhoff & Mac Lane, A survey of Modern Algebra, Section 15.6).

Things get a little more interesting when $p$ and $q$ are not irreducible. Suppose that $p=p_1^{n_1}p_2^{n_2}\ldots$. Then $F[x]/I$ is isomorphic to $\bigoplus_i F[x]/(p_i^{n_i})$. It turns out that, if $E_i=F[x]/(p_i)$, then $\bigoplus_i F[x]/(p_i^{n_i})$ is isomorphic to $E_i[t]/(t^{n_i})$ (this result is well-known, if not well-documented; a proof appears [here (see Theorem A.19)][1]. Thus for finite fields the general answer to you question is the following:

> $F[x]/I$ and $F[x]/J$ are isomorphic if and only if the numerical invariants of the polynomials $p$ and $q$ are the same, in other words, if $p=p_1^{n_1}p_2^{n_2}\ldots p_r^{n_r}$ and $q=q_1^{m_1}q_2^{m_2}\ldots q_s^{m_s}$ are the decompositions of $p$ and $q$ into irreducible factors, then there is a bijection $w:\{1,\ldots,r\}\to \{1,\ldots,s\}$ such that the degree of $q_{w(i)}$ is the same as the degree of $p_i$ and $m_{w(i)}=n_i$ for all $i\in \{1,\ldots,r\}$.


  [1]: http://www.imsc.res.in/~amri/html_notes/notesap1.html#x8-32000A.4