Is the ratio of the side and at least one diagonal of a rhombus always irrational?

No, you can make a rhombus out of four identical Pythagorean right triangles, such as (3, 4, 5).

If $2\alpha$ is the one of the angles in the rhombus and we take the side as the unit of measure, then the ratios you're interested in are $\sin\alpha$ and $\cos\alpha$.

Can they be both rational? Note that $$ \sin\alpha=\frac{2\tan(\alpha/2)}{1+\tan^2(\alpha/2)}, \qquad \cos\alpha=\frac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)}, \qquad \tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha} $$ so that $\sin\alpha$ and $\cos\alpha$ are both rational if and only if $\tan(\alpha/2)$ is rational.

Since $\alpha$ can be any angle satisfying $0<\alpha<\pi/2$, $\tan(\alpha/2)$ can assume any value between $0$ and $1$, among which there are infinitely many rational numbers.

You can try and prove that choices of $\alpha=2\arctan r$, where $0<r<1$ and $r$ is rational, are in one-to-one correspondence with the primitive Pythagorean triples.