I have given this some thought and realised that the answer to my question is yes. In case anyone comes across this question and is interested, I will post my answer here. To begin with I will make the following claim (which is essentially the crux of the result, but I will postpone a proof in order to show how the claim solves my problem first). Here is the claim:
Given any non-decreasing function $\nu : (0,\infty) \rightarrow (0,\infty)$, there exists a function $F_\nu : [0,1] \rightarrow [0,1]$ satisfying the following properties:
- $F_\nu$ is continuous, bijective and strictly increasing
- For almost-all $x \in [0,1]$ we have that $\frac{F_\nu(x+h) - F_\nu(x)}{\nu(|h|)} \rightarrow 0$ as $h \rightarrow 0$
So assuming this claim is true for now, here is a proof to the main question. To start let $J : [0,1] \rightarrow \mathbb{R}^2$ be any Osgood curve. By the Heine–Cantor theorem, $J$ is uniformly continuous and so it has some modulus of continuity $\omega : (0,\infty) \rightarrow (0,\infty)$. Without loss of generality we may assume that $\omega$ is non-decreasing and bijective.
Define $\nu : (0,\infty) \rightarrow (0,\infty)$ via $\nu(x) := \omega^{-1}(x^2)$. Now use the above claim to construct $F_\nu$ and let $A$ be the set of points satifying the second property. Then for all $x \in A$ we have that (for small enough $h$):
$$||J(F_\nu(x+h)) - J(F_\nu(x))|| \leq \omega(|F_\nu(x+h) - F_\nu(x)|) \leq \omega(\nu(|h|)) = h^2$$
Therefore $J \circ F_\nu$ is an almost-everywhere differentiable reparameterization of $J$ which retains the properties of being an Osgood curve (namely being continuous and injective).
This leaves proving the initial claim. To begin, define the following set:
$$U_q := \bigcup_{k=0}^{q-1} \bigcup_{j=0}^{3^k - 1} (\frac{3j+1}{3^{k+1}} + \frac{1}{2} \frac{1}{3^q},\frac{3j+2}{3^{k+1}} - \frac{1}{2} \frac{1}{3^q})$$
This is a collection of points in $[0,1]$ which happen to all be a distance of at least $\frac{1}{2} \frac{1}{3^q}$ away from the Cantor set. The measure of $U_q$ is given by:
$$\sum_{k=0}^{q-1}2^k(\frac{1}{3^{k+1}} - \frac{1}{3^q}) = 1 + (\frac{1}{3})^q - 2 (\frac{2}{3})^q$$
Define $C_{\frac{p}{3^n}} := \frac{p}{3^n} + \frac{1}{3^n} C$ (where $C$ is the Cantor set) and $U_{q, \frac{p}{3^n}} := \frac{p}{3^n} + \frac{1}{3^n} U_q$ to be rescaled and translated copies of the Cantor set and $U_q$ respectively.
If we take $U_q^* := \bigcap_{n=0}^{\infty} \bigcap_{p=0}^{3^n-1} U_{q(n+1), \frac{p}{3^n}}$, we find that it must have measure at least:
$$1 - \sum_{n=0}^{\infty} [2 (\frac{2}{3})^{q(n+1)} - (\frac{1}{3})^{q(n+1)}]$$
which is a quantity that tends to $1$ as $q \rightarrow \infty$.
Now let $\phi : [0,1] \rightarrow [0,1]$ be the Cantor function, and let $\phi_\frac{p}{3^n}$ be a transformation of it so that the entire staircase sits in the interval $[\frac{p}{3^n},\frac{p+1}{3^n}]$ (and define the output to be $0$ and $1$ on the left and right of this interval respectively).
What we have done so far allows us to make the following observation:
For all $x \in U_q^*$, we have that $\phi_\frac{p}{3^n}$ is constant along the interval $(x-\delta,x+\delta)$, provided that $\delta < \frac{1}{2} \frac{1}{3^{q(n+1)}} \frac{1}{3^n}$.
Define $A_n := \min(\frac{1}{n} \nu(\frac{1}{2} \frac{1}{3^{n(n+1)}} \frac{1}{3^n}),1)$, and $f_\nu : [0,1] \rightarrow \mathbb{R}$ via $f_\nu := \sum_{n=0}^{\infty} \sum_{p=0}^{3^n - 1} \frac{1}{6^n} A_{n+1} \phi_\frac{p}{3^n}$. The continuity of this function follows from the Weierstrass M-test, and strictly increasing follows from every interval containing at least one entire Cantor staircase (with non-zero height).
Now we have the following:
If $x \in U_q^*$:
then as long as $m \geq q$, we find that for all $h$ with $\frac{1}{2} \frac{1}{3^{(m+1)(m+2)}} \frac{1}{3^{m+1}} \leq |h| \leq \frac{1}{2} \frac{1}{3^{m(m+1)}} \frac{1}{3^m}$ we have:
$|f_\nu(x+h) - f_\nu(x)| \leq \sum_{n=m}^{\infty} \sum_{p=0}^{3^n - 1} \frac{1}{6^n} A_{n+1} = \sum_{n=m}^{\infty} \frac{1}{2^n} A_{n+1} \leq A_{m+1} \leq \frac{1}{m+1} \nu(|h|)$
Hence $\frac{f_\nu(x+h) - f_\nu(x)}{\nu(|h|)} \rightarrow 0$ as $h \rightarrow 0$.
Hence $\frac{f_\nu(x+h) - f_\nu(x)}{\nu(|h|)} \rightarrow 0$ as $h \rightarrow 0$, for all $x \in \bigcup_{q=1}^\infty U_q^*$. But this union has a measure zero compliment, which essentially gives us the claim.
The only thing missing is that the claim states that the image of the function should be $[0,1]$. To fix this we finally define $F_\nu := \frac{1}{f_\nu(1)} f_\nu$.