Your conjecture is true. Here is a proof.
Define $\operatorname{Av}_n(w)$ to be the number of binary words of length $n$ which avoid the pattern $w$.
Let $u$ and $v$ be binary words with $|u| = k$ and $|v|=m$ with $k < m$. We will show that $\operatorname{Av}_n(u) < \operatorname{Av}_n(v)$ for all $n$.
Define the special words $$M_n = \overbrace{00\cdots00}^n\qquad\text{and}\qquad L_n = \overbrace{00\cdots01}^n.$$
We can show fairly easily using the cluster method of Goulden and Jackson (and other ways as well, though the cluster method works easily for any pattern) that words avoiding $M_n$ and words avoiding $L_n$ have the generating functions $$m_n(x) = \sum_{r \geq 0}\operatorname{Av}_r(M_n)x^r = \frac{1-x^n}{1-2x+x^{n+1}}$$ and $$\ell_n(x) = \sum_{r\geq 0} \operatorname{Av}_r(L_n)x^r = \frac{1}{1-2x+x^n}.$$
Moreover, of all words $w$ with $|w|=n$, $M_n$ is the most avoided word and $L_n$ is the least avoided word. Formally, for $w$ with $|w|=n$ and all $r$ $$\operatorname{Av}_r(L_n) \leq \operatorname{Av}_r(w) \leq \operatorname{Av}_r(M_n).$$
This can be seen probabilistically by observing that the number of occurrences of a pattern of length $n$ in all words of length $r$ is independent of what the pattern is. Since $M_n$ "packs" the most easily (i.e., has a lot of overlaps) and $L_n$ does not "pack" at all (i.e., cannot overlap itself), it follows that $M_n$ appears as a pattern in less words overall than any other pattern and $L_n$ appears as a pattern in more words overall than any other pattern.
It should also be obvious that $\operatorname{Av}_r(L_n) \leq \operatorname{Av}_r(L_{n+1})$ for all $r$.
We need to prove one more fact: $\operatorname{Av}_r(M_{s-1}) < \operatorname{Av}_r(L_s)$ for all $r \geq s-1$. We can do this by looking at the generating functions given above. It's clear that the inequality is eventually true because the generating functions tell us that exponential growth rate of words avoiding $M_{s-1}$ is strictly less than the exponential growth rate of words avoiding $L_s$. However, we can actually prove this inequality for all $r \geq s-1$ by subtracting $m_{s-1}(x)$ from $\ell_s(x)$ and showing that the result has positive coefficients.
\begin{align*} \ell_s(x) - m_{s-1}(x) &= \frac{1}{1-2x+x^s} - \frac{1-x^{s-1}}{1-2x+x^s}\\[6pt] &= \frac{x^{s-1}}{1-2x+x^s}\\[6pt] &= x^{s-1} \cdot \ell_{s}(x). \end{align*}
Since $\ell_{s}(x)$ has positive coefficients, our difference has positive coefficients for all powers of $x$ at least $s-1$. This proves the inequality.
We now combine all of our results: for $r \geq k$ $$\operatorname{Av}_r(u) \leq \operatorname{Av}_r(M_k) < \operatorname{Av}_r(L_{k+1}) \leq \operatorname{Av}_r(L_m) \leq \operatorname{Av}_r(v).\;\;\square$$