Answer: The triples $(k, m, n)$ must satisfy $n \leq m$ and at least one of the two conditions:
$$1^{\circ}\ k \mid m-n,$$
$$2^{\circ}\ k \mid m+n \text{ and } r+n \leq m, \text{ where } r \text{ is the remainder of } m \text{ modulo } k.$$
We start from proving that if $(k, m, n)$ are as above, then the cut is possible. We identify the large square with $[0, m] \times [0, m]$. If $k \mid m-n$, then we first cut the $m \times m$ square into the square $[0, n] \times [0, n]$ and rectangles $[0, n] \times [n, m]$ and $[n, m] \times [0, m]$. These rectangles can trivially be cut into rectangles of dimension $1 \times k$. If $k \mid m+n$ and $r+n \leq m$, then we cut $[0, m] \times [0, m]$ into a square $[r, r+n] \times [r, r+n]$ and

rectangles $[0, r] \times [0, n+r]$, $[r, m] \times [0, r]$, $[r+n, m] \times [r, m]$ and $[0, n+r] \times [r+n, m]$ (they are well defined if $r+n \le m$), each of which can be trivially cut into rectangles $1 \times k$. Note that we have used the assumption $r+n \le m$ here.
Now we show that the conditions on $(k, m, n)$ are necessary. Suppose the smaller square is equal to $[p, p+n] \times [q, q+n]$; by symmetry we may assume that $q \ge 1$.
First we show that $r+n \le m$. Each unit square $[i, i+1] \times [0, 1]$, where $i \in \{p, p+1, \dots, p+n-1\}$, is contained in some rectangle $1 \times k$ coming from the cut. If we had $r+n > m$, then we would have $m-n < k$ and this would imply that each such rectangle would be "level", i.e., of the form $[t, t+k] \times [0, 1]$. Let $S$ denote the union of these "level" rectangles and let $P(S)$ denote the area of $S$. Note that we have $n \le P(S) \le m$ and $P(S)$ is divisible by $k$. This contradicts $r+n > m$.
Now we show that $k \mid m-n$ or $k \mid m+n$. Suppose that this is not true. The idea is to write an integer $a_{ij}$ in each unit square $[i-1, i] \times [j-1, j]$, $1 \le i, j \le m$, in such a way that
(a) for any rectangle $1 \times k$ of the cut the sum of numbers lying inside equals 0,
(b) the sum of all the numbers and the sum of the numbers lying inside the square $n \times n$ are different.
The existence of such a sequence clearly yields the claim.
A second idea is to work with the sequences $(a_{ij})$ of the form $a_{ij} = a_i b_j$, where $(a_i)_{i=1}^m$ and $(b_j)_{j=1}^n$ are $k$-periodic and
$$\sum_{i=1}^{k} a_i = \sum_{j=1}^{k} b_j = 0. \qquad (1)$$
This implies (a); furthermore, (b) takes form
$$\sum_{i=1}^{m} a_i \cdot \sum_{j=1}^{m} b_j \neq \sum_{i=p+1}^{p+n} a_i \cdot \sum_{j=q+1}^{q+n} b_j,$$
or, by periodicity and (1),
$$\sum_{i=1}^{r} a_i \cdot \sum_{j=1}^{r} b_j \neq \sum_{i=p+1}^{p+s} a_i \cdot \sum_{j=q+1}^{q+s} b_j, \qquad (2)$$
where $s$ denotes the remainder coming from the division of $n$ by $k$.
If $r=0$, then the left-hand side is 0. Furthermore, as $k \nmid m-n$, we have $s > 0$; it suffices to take $a_{p+1} = a_{p+2} = \dots = a_{p+s} = b_{q+1} = b_{q+2} = \dots = b_{q+s} = 1$ and choose the remaining $a_i$'s and $b_j$'s so that the periodicity and (1) hold.

Suppose then, that $r > 0$. The conditions $k \nmid m-n$, $k \nmid m+n$ imply $r \neq s$, $r+s \neq k$. We set $b_1 = b_2 = \dots = b_r = 1$ and pick the remaining $b_j$'s so that periodicity and (1) hold. Let $A = \{1, 2, \dots, r\}$, $B = \{p+1, p+2, \dots, p+s\} \pmod k$ denote the sets of indices appearing in the sums involving $(a_j)$. Note that we take the set $B$ modulo $k$. Consider two cases:
i) $A \cup B \neq \{0, 1, \dots, k-1\}$. Then, as $A \neq B$ (as $r \neq s$) and $A \neq \emptyset$, we may choose $a'_i$s with $i \in A \cup B$ in such a way that
$$\sum_{i \in A} a_i = 1, \qquad \sum_{i \in B} a_i = 0$$
(for example, if $A \setminus B \neq \emptyset$, set $a_i = 0$ for all $i \in A \cup B$ except for one $i \in A \setminus B$, for which $a_i = 1$; if $A \subset B$, take $a_i = 0$ for $i \in A \cup B$ except for one $i \in A$, for which $a_i = 1$ and except for one $j \in B \setminus A$, for which $a_j = -1$) and complete the sequence $(a_i)$ so that it satisfies periodicity and (1). This completion is possible as there exists $i \in \{0, 1, \dots, k-1\}$ not covered by $A \cup B$. Then the right-hand side of (2) is 0, while the left one is not.
ii) $A \cup B = \{0, 1, \dots, k-1\}$. Then $A \not\subseteq B$ and $B \not\subseteq A$; furthermore, as $r+s \neq k$, we have $A \cap B \neq \emptyset$. Therefore, there exist $i_1 \in A \setminus B$, $i_2 \in B \setminus A$, $i_3 \in A \cap B$ and we set $a_{i_1} = 0$, $a_{i_2} = -1$, $a_{i_3} = 1$ and $a_i = 0$ for $i \in \{0, 1, 2, \dots, k-1\} \setminus \{i_1, i_2, i_3\}$. Then we have
$$\sum_{i=1}^{r} a_i = \sum_{i \in A} a_i = 1, \quad \sum_{i=p+1}^{p+s} a_i = \sum_{i \in B} a_i = 0$$
and hence the right-hand side of (2) is 0, while the left one is not.
The proof is complete.