The answer is yes. Indeed, let $X:=\boldsymbol{X}$, $H:=\boldsymbol{H}$, and $a:=\boldsymbol{\alpha}$. By approximation and compactness, without loss of generality (wlog), the matrix $H$ is nonsingular, so that the trace in question is $$\sum_{i=1}^n Var\,l_i (X),$$ where the $l_i$'s are linearly independent linear functionals determined by the matrix $H$. By compactness, the maximum of this trace over all random vectors $X$ with mean $EX=a$ and $P(X\in[0,1]^n)=1$ is attained at some maximizing $X$. In what follows, let $X$ be such a maximizer.

Suppose that there is some point $x\in S_X\cap[0,1]^n\setminus\{0,1\}^n$, where $S_X$ is the support of the distribution $\mu_X$ of the random vector $X$. Then (i) $\{x-h,x+h\}\subset[0,1]^n$ for some nonzero vector $h$ and (ii) $\mu_X(U_x)>0$ for some set $U_x$ relatively open in $[0,1]^n$ and such that $x\in U_x$. Wlog, the length of the vector $h$ and the diameter of the set $U_x$ are small enough so that $U_{x-h}\cup U_{x+h}\subseteq[0,1]^n$. Moving now half of the mass $\mu_X(U_x)$ by the parallel translation by vector $h$ and moving the remaining half of the mass $\mu_X(U_x)$ by the parallel translation by vector $-h$, we obtain a new probability distribution of some random vector $Y$ with values in $[0,1]^n$ such that $EY=a=EX$ and $Var\,l_i (Y)\ge Var\,l_i (X)$ for all $i$, with at least one of these inequalities being strict (namely, strict for all $i$ with $l_i(h)\ne0$). This contradicts the assumption that $X$ is a maximizer. $\quad\Box$

The answer is yes. Indeed, let $X:=\boldsymbol{X}$, $H:=\boldsymbol{H}$, and $a:=\boldsymbol{\alpha}$. By approximation and compactness, without loss of generality (wlog), the matrix $H$ is nonsingular, so that the trace in question is $$\sum_{i=1}^n Var\,l_i (X),$$ where the $l_i$'s are linearly independent linear functionals determined by the matrix $H$. By compactness, the maximum of this trace over all random vectors $X$ with mean $EX=a$ and $P(X\in[0,1]^n)=1$ is attained at some maximizing $X$. In what follows, let $X$ be such a maximizer.

To obtain a contradiction, suppose that there is some point $x\in S_X\cap[0,1]^n\setminus\{0,1\}^n$, where $S_X$ is the support of the distribution $\mu_X$ of the random vector $X$. Then (i) $\{x-h,x+h\}\subset[0,1]^n$ for some nonzero vector $h$ and (ii) $\mu_X(U_x)>0$ for some set $U_x$ relatively open in $[0,1]^n$ and such that $x\in U_x$. Wlog, the length of the vector $h$ and the diameter of the set $U_x$ are small enough so that $U_{x-h}\cup U_{x+h}\subseteq[0,1]^n$. Moving now half of the mass $\mu_X(U_x)$ by the parallel translation by vector $h$ and moving the remaining half of the mass $\mu_X(U_x)$ by the parallel translation by vector $-h$, we obtain a new probability distribution of some random vector $Y$ with values in $[0,1]^n$ such that $EY=a=EX$ and $Var\,l_i (Y)\ge Var\,l_i (X)$ for all $i$, with at least one of these inequalities being strict (namely, strict for all $i$ with $l_i(h)\ne0$). This contradicts the assumption that $X$ is a maximizer. $\quad\Box$