I am familiar with this conjecture, specifically for cusp forms.
It is known that every cusp form of a congruence subgroup of $SL_2(\mathbb{Z})$ with algebraic Fourier coefficients has bounded denominators. This is because we can decompose the space of weight $k$ cuspforms of a congruence subgroup into eigenforms (newforms and pushups of newforms), which have algebraically integral Fourier coefficients. Then any form with algebraic Fourier coefficients is just a linear combination of these eigenforms over algebraic numbers, and so must have bounded denominators.
There are several examples of noncongruence cusp forms which are $n$th roots of congruence cusp forms, which means their expansions have arbitrarily large powers of $n$ in the denominator. But there is no general method yet for showing that every genuinely noncongruence cusp form has unbounded denominators.