Not an answer but rather a long comment.  Where you write

> I tried to use the classical estimation for the ratio of Gamma
> function
> $$\left|\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}\right|\leq\frac{1}{\left|\rho\right|^{k}}$$
> but since $0<k<1$ it does not work.

in fact, Stirling's Formula tells us that with $0<\beta<1$ and $\gamma\to+\infty$,
$$\left|\frac{\Gamma\left(\beta+i\gamma\right)}{\Gamma\left(\beta+k+i\gamma\right)}\right|\sim\frac{1}{\gamma^{k}}$$. 

So even assuming the Riemann Hypothesis, your series fails to converge absolutely by the Limit Comparison Test, since $$\sum_\rho\frac{1}{\gamma^k}$$ diverges for your range of $k$.  Why do you think the series might converge even conditionally for any particular value of $x$?

It's not clear to me what kind of answer you're hoping for; you say both 'uniform estimate' and also 'computing the sum.'  I don't think there's going to be any nice answer.  For $k>1$, the sum
$$\sum_{n\leq x}\Lambda\left(n\right)\left(1-\frac{n}{x}\right)^{k-1}$$ behaves nicely in that for large $n<x$, $1-n/x$ is close to $0$ and so is $(1-n/x)^{k-1}$.  But for $k<1$, $(1-n/x)^{k-1}$ is large; you are no longer truncating the sum to be continuous in $x$.