= Solution
\b[No.] For sufficiently smooth curves,
$$
Q(\gamma,s)=\kappa_\gamma'(s)
$$
is unchanged by proper Euclidean motions and transforms correctly under translations of the arc-length parameter, so it is a <pointwise Euclidean invariant of a curve>. It is not determined by the two numbers $\kappa(s)$ and $\tau(s)$.
For a concrete comparison near $s=0$, use the <plane curve reconstructed from curvature> construction with
$$
\kappa_1(s)=1,
\qquad
\kappa_2(s)=1+s.
$$
The resulting unit-speed planar curves both have $\kappa(0)=1$ and $\tau(0)=0$, but their curvature derivatives at zero are respectively $0$ and $1$. Thus no single function $f$ can satisfy $Q(\gamma,s)=f(\kappa(s),\tau(s))$ for all curves.
Indeed, the covariance conditions alone even permit nonlocal examples such as $Q(\gamma,s)=\kappa_\gamma(s+1)$. This is the distinction recorded by <pointwise Euclidean invariants need not depend only on current curvature and torsion>.
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