Your equation is incorrect, the correct condition is
$$\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2 = \text{constant}$$
I will only show the $\Leftarrow$ part here.
Let $s$ be the arc length parametrization and $\vec{t}(s), \vec{n}(s), \vec{b}(s)$ be the
vectors appear in Frenet Serret equations. Define
$$\vec{\beta}(s) = \vec{\alpha}(s) + \frac{1}{\kappa(s)}\vec{n}(s) - \frac{\dot{\kappa}(s)}{\tau(s)\kappa(s)^2} \vec{b}(s)\tag{*1}$$
Differentiate it with respect to $s$, we get:
$$\begin{align}
\frac{d}{ds}\vec{\beta}(s)
= & \vec{t} - \frac{\dot{\kappa}}{\kappa^2}\vec{n} + \frac{1}{\kappa}(-\kappa \vec{t} + \tau \vec{b} )
- \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\vec{b} -\frac{\dot{\kappa}}{\tau\kappa^2}(-\tau\vec{n})\\
= & \left(\frac{\tau}{\kappa} -\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\right) \vec{b}\\
= & \frac{\tau\kappa^2}{\dot{\kappa}}\left(\frac{\dot{\kappa}}{\kappa^3} - \frac{\dot{\kappa}}{\tau\kappa^2}\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right) \right) \vec{b}\\
= & -\frac{\tau\kappa^2}{2\dot{\kappa}}\frac{d}{ds}\left(\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2\right) \vec{b}\\
= & \vec{0}
\end{align}$$
This implies $\vec{\beta}(s) = \vec{\beta}(0)$ is a constant. From this, we get
$$
\vec{\alpha} - \vec{\beta}(0)
= -\frac{1}{\kappa}\vec{n} + \frac{\dot{\kappa}}{\tau\kappa^2} \vec{b}
\quad\implies\quad \left|\vec{\alpha} - \vec{\beta}(0)\right|^2
= \frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2
= \text{constant}.
$$
i.e $\vec{\alpha}(s)$ lies on a sphere with $\beta(0)$ as center.
Motivation of above proof
You may wonder how can anyone figure out the magic formula in $(*1)$. If you work out the
$\Rightarrow$ part of the proof where $\alpha(s)$ lies on a sphere centered at $\vec{c}$,
you should obtain a bunch of dot products between $\vec{\alpha}(s) - \vec{c}$ and $\vec{t}(s)$, $\vec{n}(s)$ and $\vec{b}(s)$. In particular, you should get:
$$\begin{cases}
\vec{t} \cdot (\vec{\alpha} - \vec{c}) & = 0\\
\vec{n} \cdot (\vec{\alpha} - \vec{c}) & = -\frac{1}{\kappa}\\
\vec{b} \cdot (\vec{\alpha} - \vec{c}) &= \frac{\dot{\kappa}}{\tau\kappa^2}
\end{cases}$$
Using these, you can express the center $\vec{c}$ in terms of $\kappa, \tau$ like what we have in $(*1)$. If the curve does lie on a sphere, then the "center" should not move as $s$ changes. The proof of the $\Leftarrow$ part above is really using the given condition to verify the "center" so defined doesn't move.
Suppose $\alpha$ parametrized by the arc length, remember that
$$s(t)=\int_0^t|\alpha'(s)|\,ds,$$
hence $s'(t)=|\alpha'(t)|=1$.
Now
$$
\begin{matrix}
\langle\gamma'(t)-\alpha(0),u\rangle&=&\langle\alpha'(t)-s'(t)\cos(\theta) u-\alpha(0),u\rangle\\
&=&\langle\alpha'(t),u\rangle-\langle s'(t)\cos(\theta)u,u\rangle-\langle\alpha(0),u\rangle\\
&=&\cos(\theta)-\cos(\theta)\langle u,u\rangle-0=0.
\end{matrix}
$$
We use $s'(t)=1$ and the fact that the vector $\alpha(0)$ lies in the plane ortogonal to $u$.
Best Answer
Suppose that $C=(a,b,c)$ is the center of the sphere on which $\alpha$ is supposed to lie. The square of the distance $d(C,\alpha)$ has to be constant. Therefore by differentiation
$$\langle \alpha^\prime(t), \alpha(t) - C \rangle$$ has to be equal to zero for all $t \in \mathbb R$. I.e.
$$\begin{aligned} 2\sin t \cos t(a - r\sin^2 t) + \cos t(b -r \sin t) - \sin t(c - r \cos t)&=\\ 2\sin t \cos t(a - r\sin^2 t) + b \cos t - c\sin t=0 \end{aligned}$$
Using $t = 0$ we get $b=0$ and $c =0$ follows from plugging in $t= \pi/2$. We're left with the equation $$\sin t \cos t(a - r\sin^2 t) = 0$$ which should be valid for all $t \in \mathbb R$. And that can't be as it would imply $a - r\sin^2 t=0$ for all $t\in (0,\pi/2)$.
Therefore, the curve doesn't lie on a sphere.