The answer is no: if $\varepsilon$ is small enough, then for every open $A \subset [0,1]^2$ of measure at least $1/100$, there exists a smooth curve $\gamma$ of length $\leq 1$ such that $\gamma+A$ contains a $\varepsilon$-ball. The idea is to randomly shift by translates first at extremely small scales to cover a lot of balls (or squares) at that scale, then at less small scales to cover a lot of balls (or squares) at this larger scale, and so forth until one is able to cover a macroscopic ball or square.

First some reductions. By enlarging $\varepsilon$ a little bit, we can assume it is a negative power of two, and replace "$\varepsilon$-ball" by "dyadic $\varepsilon$-square". By inner regularity, $A$ contains a finite union of balls of measure at least $1/200$, hence contains a finite union $A'$ of closed dyadic squares of measure at least $1/400$. If $N$ is large enough, we can view $A'$ as a union of dyadic $2^{-N}$-squares, and assume that the $2^{-N}$-neighbourhood of $A'$ lies in $A$. It will then suffice to find a piecewise polygonal path $\gamma'$ of length at most $1/2$ such that $\gamma'+A'$ contains a dyadic $\varepsilon$-square, since one can smooth $\gamma'$ out at a scale much less than $2^{-N}$ to find a smooth curve $\gamma$ of length at most $1$ such that $\gamma+A \subset \gamma'+A'$.

Now introduce the hyperdyadic scales $\varepsilon_n := \varepsilon^{2^n}$ and integers $J_n := \lfloor \log^{100} \frac{1}{\varepsilon_n} \rfloor$. Clearly there is $n_0$ such that $\varepsilon_{n_0+1} \leq 2^{-N}$, hence $A'$ is now a union of dyadic $\varepsilon_{n_0+1}$-squares. We will choose, for each $n=0,\dots,n_0$, a set $H_n$ of $J_n$ shifts in the ball $B(0, 10 \varepsilon_n)$. By an easy induction on $n_0$, it is possible to find a piecewise polygonal path $\gamma'$ passing through the sumset
$$ H_0 + H_1 + \dots + H_{n_0}$$
(which is a finite set of cardinality at most $J_0 \dots J_{n_0}$) of length at most
$$ \sum_{n=0}^{n_0} 20 \varepsilon_n J_0 \dots J_n$$
which one can calculate to be less than $1/2$ if $\varepsilon$ is small enough (the double exponential decay of the $\varepsilon_n$ beats the quadratic exponential growth of the $J_0 \dots J_n$). Hence it will suffice to locate $H_0,\dots,H_{n_0}$ such that
$$ H_0 + H_1 + \dots + H_{n_0} + A'$$
contains a dyadic $\varepsilon$-square.

We will establish the more general claim that for any $0 \leq n \leq n_0+1$, we can select $H_n,\dots,H_{n_0}$ such that
$$ H_n + \dots + H_{n_0} + A'$$
contains a union of dyadic $\varepsilon_{n}$-squares in $[0,1]^2$, of total measure at least $\frac{1}{400} - 1000^{-n-1}$; setting $n=0$ will give the claim.

We prove this claim by downward induction on $n$. When $n=n_0+1$ the claim follows since $A'$ itself is a union of dyadic $\varepsilon_{n_0+1}$ squares in $[0,1]^2$ of measure at least $\frac{1}{400}$. Now assume inductively that $0 \leq n \leq n_0$ and the claim has already been proven for $n+1$, thus we have already located $H_{n+1},\dots,H_{n_0}$ such that
$$ H_{n+1} + \dots + H_{n_0} + A'$$
contains a union $B$ of dyadic $\varepsilon_{n+1}$-squares in $[0,1]^2$, of total measure at least $\frac{1}{400} - 1000^{-n-2}$.

We cover $B$ by dyadic $\varepsilon_n$-squares. If we let $B'$ be the union of all the dyadic $\varepsilon_n$-squares in which $B$ has relative density at least $1000^{-n-2}$, then $B \backslash B'$ has measure at most $1000^{-n-2}$, hence $B'$ (which contains $B \cap B'$) has measure at least $\frac{1}{400} - 2 \times 1000^{-n-2}$. We now use the probabilistic method, picking $H_n$ to be $J_n$ random elements of $B(0,10\varepsilon_n)$. A standard union bound calculation (discretising each $\varepsilon_n$-square at scale $\frac{1}{10} \varepsilon_{n+1}$, say into a lattice of cardinality $O( (\varepsilon_n/\varepsilon_{n+1})^2 )$) shows that each $\varepsilon_n$-square in $B'$ will lie in $H_n+B$ with probability at least
$$ 1 - O( (\varepsilon_n/\varepsilon_{n+1})^2 ( 1 - 1000^{-n-3} )^{J_n} )$$
which by the choice of parameters can be seen to be at least $1 - 1000^{-n-2}$. Thus, by linearity of expectation (first moment method), one can choose $H_n$ so that $H_n+B$ covers a subcollection of $\varepsilon_n$-squares in $B'$ of measure at least $\frac{1}{400} - 1000^{-n-1}$, closing the induction.