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 $\frac d{dt}h_{X,Y}(t)=\frac d{du}h_{X+tY,Y}(u)|_{u=0}$).
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}
and then 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 $u_+^a:=\max(0,u)^a$ and
$\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed.

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that

$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.