nt.number theory - weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

I am not that sure that your question can be answered positively, but the following is merely speculation, so you should not pin me down on it. The basic idea is that there might be "more" weight 4 forms than rigid CY 3-folds.

It seems that (but Kevin probably knows this better than me) that it is still an open question whether there are finitely or infinitely many weight 4 eigenforms up to twisting.

Suppose there were infinitely many weight 4 eigenforms up to twisting. To realize every weight 4 eigenform we then need infinitely many $overline{mathbb{Q}}$-isomorphism classes of rigid CY 3-folds defined over $mathbb{Q}$. All hodge numbers of a rigid CY 3-fold are a priori fixed, except for $h^{1,1}$ and $h^{2,2}$, which coincide. The Euler characteristic of a rigid CY 3-fold is $2h^{1,1}$ and rigid CY 3-folds do not admit deformation, hence in order to realize every weight 4-form we either find a Hodge diamond $D$ such that

{ $X | X$ smooth projective complex variety with Hodge diamand $D$}/deformations

is infinite, or the absolute value of the Euler characteristic of a CY 3-folds can be arbitrarily large. The first conclusion would be quite remarkable, the second would solve an open problem (as far as I know).

Suppose now there were only finitely many weight 4 eigenforms up to twisting.
If you want to avoid the above mentioned problems you need that every eigenform $f$ is realized by a CY 3-fold $Y_f$ admitting an involution, so that a twist of $f$ is realized by a twist of $Y_f$.

Still it is not clear whether there are enough rigid CYs to realize every eigenform. Some computational evidence can be found in the book of Christian Meyer, Modular Calabi-Yau Threefolds. He realizes close to 100 eigenforms (up to twisting). The corresponding list at http://www.fields.utoronto.ca/publications/supplements/weight4.pdf
contains much more forms. The smallest level that he could not realize is 7.

If you allow $h^{2,1}$ to be nonzero, i.e., you allow the motive of the form to be a factor of $H^3$ or if you are happy to work with quasi-projective varieties $Y$ such that its completions is a CY 3-fold then you are in a much better position.