Let $\alpha : [a,b] \rightarrow \mathbb R^3$ be a $C^1$ mapping (curve). Then $\alpha$ has a length. If $\alpha'(t)\neq 0$ for all $t\in [a,b]$ then, denoting $$ \sigma(t)=\int_a^t |\alpha'(u)|du, $$ $$ \beta(s)=\alpha(\sigma^{-1}(s)), $$ we obtain that $\beta$ is a curve in the natural parametrization ($|\beta'(s)|=1$ for all $s$).
Assume that the condition $\alpha'(t)\neq 0$ for all $t\in [a,b]$ is not satisfied. Can $\alpha$ be reparametrized in order to be in the natural parametrization?
You cannot have a smooth arclength parametrization unless the curve itself is geometrically smooth. The standard example $\alpha(t) = (t^2,t^3)$ illustrates the issue: the curve has a cusp, where no tangent line exists.
The map $\alpha$ manages to be smooth because it crawls across the cusp with vanishing speed. But if you attempt to travel along the same curve with unit speed, the map will not have a derivative as it passes the cusp: its direction of travel will abruptly change.
But you can consider $\beta=\alpha\circ \sigma^{-1}$ anyway: this composition is well-defined even when $\sigma^{-1}$ is multivalued (stretches some points into intervals), because $\alpha$ is constant on every interval where $\sigma $ is constant. By construction, $\beta$ is the unit speed parametrization in the sense that the length of each segment $\beta_{[t,s]}$ for $t<s$ is equal to $s-t$. In particular, $\beta$ is a Lipschitz map, hence differentiable a.e. At every point of differentiability, $|\beta'(t)|=1$.
So, if you are content with a Lipschitz, rather than $C^1$, parametrization, $\beta$ does what you want.