As the question is asked, the answer is "no": if a continuous curve $\gamma:\mathbb{R}_{\geq 0}\to [0,1)^2$ is self-avoiding, i.e., injective, then the image $\gamma(\mathbb{R}_{\geq 0})$ is nowhere dense in $[0,1)^2$. Indeed, the images $\gamma([0,T])$ are compact, hence closed. Also, compactness implies the inverse map $\gamma^{-1}:\gamma([0,T])\to[0,T]$ is continuous, i.e., the image $\gamma([0,T])$ is homeomorphic to an interval. Hence, it cannot contain any balls, for if it did, removing a point from it would make the fundamental group non-trivial, while removing a point from an interval cannot do so.

Baire's theorem now implies that $\gamma(\mathbb{R}_{\geq 0})=\cup_{T\in\mathbb{N}}\gamma([0,T])$ is nowhere dense.

**Update:** one way to relax the self-avoidance property: we say that a curve is non-self-crossing if it belongs to the closure of the set of injective curves with respect to the metric 
$$
d(\gamma_1,\gamma_2)=\inf_\sigma \sup_t|\gamma_1(\sigma(t))-\gamma_2(t)|.
$$
where the infimum is over all parametrizations $\sigma$. Such a curve may intersect its past trace, but then it bounces back on the same side it came from. The $SLE_\kappa$ curves are like that (but non-simple) for $\kappa>4$, and space filling for $\kappa\geq 8$. It is conjectured that self-avoiding walk on a lattice $\epsilon \mathbb{Z}^2$ (uniformly chosen from all walks staying in a domain $\Omega$ and with given endpoints) converges in the scaling limit $\epsilon\to 0$ to the $SLE_{8}$ process. See Duminil-Copin, H., Kozma, G., & Yadin, A. (2014). Supercritical self-avoiding walks are space-filling. In Annales de l'IHP Probabilités et statistiques (Vol. 50, No. 2, pp. 315-326).