Why would a non-squareness of $A$ pose a problem? If $A\in\mathbb{R}^{m\times n}$, $A_0=A$, you first compute $A_1=Q_1A_0$, where $Q_1$ is a product of Givens rotations annihilating the elements $2,\ldots,m$ of the first row. Similarly, given $A_{k-1}$ having zeroed out the "lower trapezoid" in the first $k-1$ rows, you compute $A_k=Q_kA_{k-1}$, where $Q_k$ are again Givens rotations annihilating the rows $k+1,\ldots,m$ of the column $k$ of $A_{k-1}$. You repeat the steps until $k=\min\{m,n\}$.
If $A$ has full rank and, say, $m\geq n$, this way you get a decomposition $A=QR$, where $Q\in\mathbb{R}^{m\times m}$ is orthogonal and $R\in\mathbb{R}^{m\times n}$ has the structure
$$
R=\begin{bmatrix}\tilde{R} \\ 0\end{bmatrix},
$$
where $\tilde{R}\in\mathbb{R}^{n\times n}$ is upper triangular.
Note that if $Q=[\tilde{Q},\hat{Q}]$, where $\tilde{Q}\in\mathbb{R}^{m\times n}$, then
$$
A = QR = [\tilde{Q},\hat{Q}]\begin{bmatrix}\tilde{R} \\ 0\end{bmatrix} = \tilde{Q}\tilde{R},
$$
so if you want to get the "economy" factors, simply take the first $n$ columns of $Q$ and first $n$ rows of $R$.
It's is useful to think of Givens rotations in $\mathbb{R}^2$ "plane". To make zeros in a 2-vector $[\alpha,\beta]^T$ by a rotation
$$
\begin{bmatrix}c & -s \\ s & c\end{bmatrix}\begin{bmatrix}\alpha\\\beta\end{bmatrix}=\begin{bmatrix}r \\ 0\end{bmatrix},
$$
you take $c=\alpha/r$ and $s=-\beta/r$, where $r=\sqrt{\alpha^2+\beta^2}$.
So in your example with
$$
A = \begin{bmatrix}
1 & -2 & 3 \\
1 & -3 & 5 \\
2 & 3 & 1 \\
-3 & 4 & 2\\
\end{bmatrix},
$$
it is useful to think about annihilating the bottom-left element using the third row as if you want to do the above thing on the vector [2,-3]^T.
Hence
the proper choice of $c$ and $s$ is
$$
c = 2 / r, \quad s = 3 / r, \quad r = \sqrt{2^2+(-3)^2} = \sqrt{13}.
$$
Best Answer
Yes, in Hessenbergform, you only need ~$n$ Givens-Rotations per QR step, in full form that requires ~$n^2/2$ rotations.
It is not possible to just get the triangular form. Would be nice, since that would have the eigenvalues directly on the diagonal.