The answer to this question is yes. 

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$ and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. The inequality in question is equivalent to the following: 
\begin{equation}
	h(t):=h_{X,Y}(t):=[E(X+tY)^a-E^a(X+tY)]^{1/a}-(EX^a-E^a X)^{1/a}-t(EY^a-E^a Y)^{1/a}\le0
\end{equation}
for all $t\ge0$, and hence to the condition $h'(0)\le0$ (because $\frac d{dt}h_{X,Y}(t)=\frac d{du}h_{X+(t+u)Y,Y}(u)|_{u=0}$). In turn, the condition $h'(0)\le0$ can be rewritten as 
\begin{equation}
	(EX^{a-1}Y-E^{a-1}X\,EY)(EX^a-E^a X)^{1/a-1}\le(EY^a-E^a Y)^{1/a}. 
\end{equation}
The latter inequality is trivial if $EX^{a-1}Y-E^{a-1}X\,EY\le0$. Otherwise, it can be rewritten in this H\"older-like form: 
\begin{equation}
	F(\mu):=(EX^{a-1}Y-E^{a-1}X\,EY)^a-(EX^a-E^a X)^{a-1}(EY^a-E^a Y)\le0, 
\end{equation}
where $\mu=\mu_Y$ is the probability distribution of $Y$, with random variable (r.v.) $X$ considered fixed. Obviously, the function $F$ is convex on the set $M$ of all probability distributions $\mu_Y$ on $S$ such that $EX^{a-1}Y-E^{a-1}X\,EY\ge0$. So, the maximum of $F$ on the compact convex set $M$ is attained when either $\mu$ is a Dirac measure (that is, when the r.v. $Y$ is a nonnegative constant) or at a boundary point of $M$ where $EX^{a-1}Y-E^{a-1}X\,EY=0$. 
However, if $Y$ is a nonnegative constant $c$, then $EX^{a-1}Y-E^{a-1}X\,EY=c(EX^{a-1}-E^{a-1}X)\le0$, since $1<a\le2$. So, the maximum of $F$ on the compact convex set $M$ is attained when $EX^{a-1}Y-E^{a-1}X\,EY=0$, and then clearly $F(\mu)\le0$. 
This completes the proof for the case $1\le a\le2$. 

Following the lines of this proof, it should be clear that the inequality in question fails to hold in general for any $a>2$.