$\newcommand\Ga\Gamma$Using the formula $$\frac1{(1+k)^r}=\frac1{\Ga(r)}\int_0^\infty dx\,x^{r-1}e^{-(1+k)x}$$ and then the substitution $v=-\ln(1-e^{-x})$, we get \begin{equation} s_{n,r}:=\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(1+k)^r} =\frac1{\Ga(r)}\, I_{n,r}, \end{equation} where, with $m:=n+1$, \begin{equation} I_{n,r}:=\int_0^\infty dv\,g(v)^{r-1} e^{-mv}=I_{n,r,1}+I_{n,r,2}, \end{equation} \begin{equation} I_{n,r,1}:=\int_0^{1/\sqrt m} dv\,g(v)^{r-1} e^{-mv},\quad I_{n,r,2}:=\int_{1/\sqrt m}^\infty dv\,g(v)^{r-1} e^{-mv}, \end{equation} \begin{equation} g(v):=-\ln(1-e^{-v}). \end{equation}
Note that $g$ is decreasing from $\infty$ to $0$ on $(0,\infty)$, \begin{equation} g(v)\underset{v\downarrow0}\sim-\ln v,\quad g(v)\underset{v\to\infty}\sim e^{-v}, \end{equation} \begin{equation} 0<v\le1\implies g(v)\le-\ln(c v)=|\ln c|+|\ln v|, \end{equation} where $c:=1-e^{-1}$, \begin{equation} v>0\implies g(v)\ge-\ln v. \end{equation}
It follows that $g(v)^{r-1}=O(\ln^{|r-1|}m+e^{|r-1|v})$ for $v>1$ and hence \begin{equation} I_{n,r,2}=O(e^{-\sqrt m}\ln^{|r-1|}m). \end{equation}
Next, \begin{equation} I_{n,r,1}=\frac1m\int_0^{\sqrt m} du\,g(u/m)^{r-1} e^{-u}, \end{equation} and, for $u\in(0,\sqrt m)$, \begin{equation} 0<g(u/m)^{r-1}\le(-\ln(u/m))^{r-1}\le2^{1-r}\ln^{r-1}m \end{equation} if $0<r\le1$, and \begin{equation} 0<g(u/m)^{r-1}\le(|\ln c|+|\ln(u/m)|)^{r-1}=O(|\ln c|^{r-1}+|\ln u|^{r-1}+\ln^{r-1}m) \end{equation} if $r>1$. Also, for each real $u>0$ we have $g(u/m)\sim -\ln(u/m)\sim\ln m$ (as $n\to\infty$).
So, by dominated convergence, \begin{equation} I_{n,r,1} \sim\frac1m\int_0^{\sqrt m} du\,\ln^{r-1}m\, e^{-u}\sim\frac1m\,\ln^{r-1}m \end{equation} and thus \begin{equation} s_{n,r}\sim\frac1{\Ga(r)}\frac{(\ln n)^{r-1}}n \end{equation} for each real $r>0$, just as what you got for natural $r$.
