Your hunch is wrong. A non-rectangular non-rhomboid parallelogram with integer side and diagonal lengths exists:
Suppose that the parallelogram is $ABCD$ with $AB=a$, $AD=b$ and $BD=c$. By the law of cosines:
$$\cos\angle DAB=\frac{a^2+b^2-c^2}{2ab}$$
Since $\angle CDA=\pi-\angle DAB$, $\cos\angle CDA=-\cos\angle DAB$. Then
$$AC^2=d^2=a^2+b^2-2ab\cos\angle CDA=a^2+b^2+2ab\cdot\frac{a^2+b^2-c^2}{2ab}=2a^2+2b^2-c^2$$
$$\color{red}{c^2+d^2=2(a^2+b^2)}$$
If $c=d$ we have a rectangle; if $a=b$ we have a rhombus. Thus, we look for a number that is a sum of two unequal squares $a$ and $b$ whose double is also a sum of two unequal squares $c$ and $d$, with triangle inequalities satisfied to ensure the parallelogram is non-degenerate: $|a-b|<c,d<a+b$.
Fixing $a$ and $b$, a trivial choice is $c=a-b$ and $d=a+b$ because $2(a^2+b^2)=(a-b)^2+(a+b)^2$. However, these assignments do not satisfy the triangle inequalities, making numbers that have multiple representations as sums of two squares valuable for this problem. I used OEIS A025426 to find them; the first number I saw was $145=9^2+8^2=12^2+1^2$, whose double is $290=13^2+11^2=17^2+1^2$. The first representations listed here allowed me to quickly construct the parallelogram above, although it is not the smallest: there is a parallelogram with sides 4 and 7, diagonals 7 and 9.
The numbers being squared in the above equations relate to the above parallelogram as follows. The lengths of its sides are 9 and 8, and of its diagonals 13 and 11. If its left side is extended upwards along the same straight line by 8 (so that it is now 16), the resulting quadrilateral has sides 16, 9, 8 and 11, and its diagonals are 17 and 13.
Here is a way to efficiently generate infinitely many such integer parallelograms. Let $r$ and $s$ be two coprime integers with $r>s>0$ and $(r,s)\ne(3,1)$. The product $(2^2+1^2)(r^2+s^2)$ can be written as a sum of two squares in two different ways (the Brahmagupta–Fibonacci identity):
$$(2^2+1^2)(r^2+s^2)=(2r+s)^2+(2s-r)^2=\color{blue}{(2s+r)^2+(2r-s)^2}$$
From these two representations, we can write twice the product as two different sums of two squares too:
$$2(2^2+1^2)(r^2+s^2)=\color{blue}{(2(r+s)-(r-s))^2+(2(r-s)+(r+s))^2}
=(2(r-s)-(r+s))^2+(2(r+s)^2+(r-s))^2$$
To satisfy the triangle inequalities we choose
$$\color{blue}{a=2r-s\qquad b=2s+r\qquad c=2(r+s)-(r-s)\qquad d=2(r-s)+(r+s)}$$
These are guaranteed to form a non-rectangular non-rhomboid integer parallelogram with the given restrictions on $r$ and $s$. The one pictured at the top of this answer corresponds to $(r,s)=(5,2)$ and the smallest instance (the one with side lengths 4 and 7) corresponds to $(r,s)=(3,2)$.
