Using the idea of one of the proof's of the famous Dehn's tiling problem (see for example here), assume that such a cutting of a square is possible.

Let $S$ the unit square, and $S=\cup_{j=1}^n R_j$ and $$[0,r]\times[0,1/r]=T=\cup_{j=1}^n R_j',$$ were $R'_j$ is produced be translating and/or rotating by $\pi/2$ the $R_j$, and $r$ irrational - we assume that there are no overlaps among the $R_j$'s and no overlaps among the $R_j'$'s, meaning that $R_j\cap R_k$ could only be part of an edge.

As $r$ is irrational, it is possible to extend $\{1,r\}$ to a Hamel basis $\mathcal B$ of $\mathbb R$ over $\mathbb Q$. Define a linear functional $\mathbb f :\mathbb R\to\mathbb R$ as:
$$
f(1)=1, \,\, f(r)=0, \,\, \text{and}\,\, f(b)=0, \,\,\text{for all}\,\, b\in\mathcal B
\smallsetminus\{1,r\}.
$$
Then for any rectangle $R=[a,b]\times[c,d]$ define its $f-$area as
$$
A_f(R)=f(b-a)f(d-c).
$$
It is not hard to check that $A_f$ is finitely additive, and it is invariant to rotations of $k\pi/2$, $k\in\mathbb Z$ and translations. In particular, $A_f(R_j)=A_f(R_j')$, for all $j$.
Then
$$
0=f(r)f(1/r)=A_f(T)=\sum_{j=1}^n A_f(R_j')=\sum_{j=1}^n A_f(R_j)=A_f(S)=f(1)f(1)=1,
$$
which is a contradiction. Hence such cutting is not possible.

**Note.** The use of the Axiom of Choice (i.e., the fact that the linear space $\mathbb R$ over $\mathbb Q$ possesses a Hamel basis) can be avoided, as $f$ does not have to be defined on the whole of $\mathbb R$, but on the subspace of $\mathbb R$ over $\mathbb Q$, which is spanned by the sizes of the rectangles $R_j$. It is also noteworthy, that the additivity of $A_f$ is used in the proof of Dehn's Theorem.