Quotient is principal Let $R$ be a finite commutative ring, let $J$ be a maximal ideal of $R$ and $n$ some positive integer greater or equal than $2$. Is it always true that every ideal of the quotient $R/J^{n}$ is principal?

No.  For instance if $R$ is local then $J^n = 0$ for sufficiently large $n$ and then you are just asking whether a finite, local commutative ring must be principal.  The answer is certainly not.
Examples have come up before on this site.  One natural one is $R = \mathbb{F}_p[x,y]/\langle x,y \rangle^2$.  This is somehow especially instructive: the maximal ideal $\mathfrak{m} = \langle x,y \rangle$ is not monogenic since $\operatorname{dim}_{R/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 = \operatorname{dim}_{\mathbb{F}_p} \mathfrak{m} = 2$.  I felt I understood Zariski tangent spaces a bit better after thinking about this example.
Another simple example is $\mathbb{Z}[\sqrt{-3}]/\langle 2 \rangle$.  More generally, if $R$ is any nonmaximal order in a ring of integers in a number field, then there will be some ideal $I$ of $R$ such that $R/I$ is finite and nonprincipal.
Added: Over the summer I had the occasion to think about finite non/principal rings.  For more information about when finite local rings are principal, and especially the connection to residue rings of orders in number fields, see $\S$ 1 of this paper.
