The smallest possible area is $35$.

Let $a$ be the length of the smallest square $A$. If $A$ touches a side but not a corner, then it is surrounded by two larger squares $B$ and $C$. Then there is a gap between $B$ and $C$ with length $a$. It is not possible to place some larger squares in this gap, contradiction. Similarly, the smallest square cannot be placed at a corner.

The distance from $A$ to a side is at least $a+1$. Therefore, each side has length at least
$$(a+1) \times 2 + a = 3a + 2 \ge 5.$$

If both side lengths are at least $6$, then the area is at least $36$. Thus, we may assume there is a side with length $5$, and $a=1$.
Since there is no square of side length $1$ touching this side, either there is a $5 \times 5$ square, or a $3 \times 3$ and a $2 \times 2$ squares touching this side. The former case can be ignored since we can remove such a $5 \times 5$ square. Now, suppose there is a $3 \times 3$ square $B$ and a $2 \times 2$ square $C$ as shown. Then $A$ cannot be placed in the first $3$ columns (or there is a gap of length $1$). The same holds for the opposite side. Therefore, the other side length of the rectangle is at least $3+1+3=7$. This shows the area is at least $5 \times 7 = 35$.

An example of a $5 \times 7$ rectangle is given below.