No, from experimentation point of view (check Emilio's comments). Yes, from the "practical"-theoretical point of view. No, from the rigorous theoretical point of view. (edited paragraph).
Let's assume a planet with mass $m_p$, and a object in which will make a slingshot, of mass $m$. Planet has speed $v_p$. Mass $m$ has speed $v$. After slingshot, mass $m$ has speed $2v_p + v$.
Linear momentum must conserve. So, initially $p_i$ and after slingshot $p_f$. $$ p_i = mv + m_p v_p = m(v + 2v_p) + m_p v_p' = p_f $$
Where $v_p'$ is the final velocity of the planet. We can isolate it: $$ v_p' = \frac{mv + m_p v_p - m(v + 2v_p)}{m_p} = \frac{m_p v_p - 2mv_p}{m_p} = v_p - \frac{2m}{m_p}v_p $$
Therefore, the variation of planet speed $\Delta v_p = v_p' - v_p$ is: $$ \Delta v_p = -\frac{2m}{m_p}v_p $$
Now we can throw up $n$ times a mass $m$ object to peform slingshot, which means, $n$ slingshots. If $m_p >> m$ (which is of course true since you won't slingshot a planet in another planet) it is valid the approximation such that $\Delta v_p \approx dv_p$ and then we integrate over $n$ slingshots. $$ \frac{dv_p}{v_p} = -\frac{2m}{m_p}dn \quad\Longrightarrow\quad \int \frac{dv_p}{v_p}dn = -\int \frac{2m}{m_p}dn $$
$$ \ln v_p = \frac{2m}{m_p}n + C $$
We can find out the integral constant such that be in function of $v_0$, where $v_0$ is the initial speed of the planet. Then we get a function $v_p(n)$, which means, velocity of the planet is dependent from the amount $n$ of slingshots performed.
We end up with: $$ v_p(n) = v_0 \exp\left(-\frac{2m}{m_p}n\right) $$
Where $n$ is the number of slingshots. So, you can see each slingshot the planet speed drops exponentially, and therefore, rigorously never reaches zero. But, it will be close enough to zero after a lot slingshots. A nice observation, from here we can notice that after a slingshot, the planet speed is independent from the initial speed of mass $m$ object. Only depends on mass $m$ of the object.