I've supplied an answer to another question here https://mathematica.stackexchange.com/questions/222064/line-integration-given-tangent-vector/222205#222205
I'd like somebody more familiar with vector calculus to check as I'm not 100% confident. We have a smooth curve $C$ where $\alpha:[0,1]\rightarrow\mathbb{R}^3$. Suppose the tangent vectors of $C$ are always of unit length so $||\alpha'(t)||_2=1$. Typically these would be lines or circles or helices, but what other curves might we expect?
And suppose we have a vector field $\mathbf{F}$ such that $\mathbf{F}[\alpha_x(t),\alpha_y(t),\alpha_z(t)]=\alpha^\prime(t)$ - that is the field $\mathbf{F}$ is exactly the unit tangent vector when restricted to the curve $C$.
Then we have the following line integral along $C$:
$$ \int_{C}\nabla\cdot\mathbf{F}\ ds = \int_{a}^{b}\nabla\cdot\alpha^\prime(t)||\alpha'(t)||_2 \ dt=\int_{a}^{b}\nabla\cdot\alpha^\prime(t)\ dt $$
I claim this equals $0$ as there should be no average net flux in/out of an infinitesimal box that travels along the curve. At any point, a unit vector enters and exits the box. Are the next steps below valid or do I need to use the chain-rule on the partial derivatives for substituting $x=\alpha_x(t),y=\alpha_y(t),x=\alpha_z(t)$ ?
$$=\int_{a}^{b}\frac{\partial}{\partial x}\frac{d \alpha_x}{dt}+\frac{\partial}{\partial y}\frac{d \alpha_y}{dt}+\frac{\partial}{\partial z}\frac{d \alpha_z}{dt}dt$$
$$=\frac{d}{d x}\int_{a}^{b}\frac{d \alpha_x}{dt}dt+\frac{d}{d y}\int_{a}^{b}\frac{d \alpha_y}{dt}dt+\frac{d}{d z}\int_{a}^{b}\frac{d \alpha_z}{dt}dt=0+0+0$$
No, the closed-loop integral of a divergence is not always zero (it is only the closed-loop integral of the gradient of a scalar function which is always $0$). Here's a simple counter example. Let $F: \Bbb{R}^3 \to \Bbb{R}^3$ be defined by \begin{align} F(x,y,z) = (-y,x,z), \end{align} and consider the curve $\alpha: [0,2\pi] \to \Bbb{R}^3$ defined by \begin{align} \alpha(t) = (\cos t, \sin t, 0). \end{align} So, the curve $C$ parametrized by $\alpha$ is the unit circle in the $x$-$y$ plane. It is clear that \begin{align} (F \circ \alpha)(t) &= (-\sin t, \cos t, 0) = \alpha'(t), \end{align} and that $\lVert \alpha'(t) \rVert = 1$. So, $F$ and $\alpha$ satisfy all of your hypotheses. However, the divergence of $F$ is $1$. So, \begin{align} \int_C (\nabla \cdot F)\, ds &= \int_C \, ds = \int_0^{2\pi} \lVert \alpha'(t)\rVert\, dt = 2\pi \neq 0. \end{align}
The biggest flaw in your argument (which I didn't really understand) is that you started talking about the divergence of $\alpha'$, i.e $\nabla \cdot (\alpha')$. This makes no sense, because the divergence is only defined for vector fields which are defined on open subsets of $\Bbb{R}^3$ (i.e for functions of $3$-variables). However, $\alpha'$ is simply a map $[0,2\pi] \to \Bbb{R}^3$, which is a function of one-variable. So, things like $\dfrac{\partial \alpha_1}{\partial x}, \dfrac{\partial \alpha_2}{\partial y}$ etc are not even well-defined objects.
Notice that the correct statement is $F \circ \alpha = \alpha'$, NOT $F = \alpha'$. After all, how can a function of $3$-variables be equal to a function of $1$ variable? This is why the step in going from $\nabla \cdot F = \nabla \cdot \alpha'$ is completely wrong; actually it's not just wrong, the RHS isn't even well-defined.
Hopefully this clarifies the situation.