No. E.g. when $D'=\operatorname{diag}(2,1),\,D=I_2$ and $A=\pmatrix{1&2\\ 2&5}$, the difference
$$
D'AD'-DAD=\pmatrix{4&4\\ 4&5}-\pmatrix{1&2\\ 2&5}=\pmatrix{3&2\\ 2&0}
$$
is indefinite.
The largest number of elements that a collection of orthogonal positive semidefinite nonzero $N\times N$ matrices can have is $N$.
First note the following facts. For two positive semidefinite $N\times N$ matrices $A$ and $B$, one has the following equivalence:
$$ AB=0 \quad\iff\quad \langle A,B\rangle=0.$$
Moreover, for positive semidefinite matrices $A$ and $B$, it holds that $AB=0$ if and only if $BA=0$. Thus, $\langle A,B\rangle =0$ implies that $AB-BA=0$, hence the matrices $A$ and $B$ commute.
Now, if $\{A_1,A_2,\dots,A_m\}$ is any collection of pairwise orthogonal, positive semidefinite matrices (i.e., $\langle A_i,A_j\rangle =0$ for all $i,j\in\{1,2,\dots,m\}$ such that $i\neq j$), it must be the case that they are all simultaneously commuting. Because they are pairwise commuting, this collection is simultaneously diagonalizable. Thus, we may suppose without loss of generality that $A_1,A_2,\dots,A_m$ are all diagonal matrices. The dimension of the span of all diagonal matrices is $N$, so the span of $\{A_1,\dots,A_m\}$ is at most $N$.
Best Answer
The statement is false. Consider $$A=\begin{bmatrix} 6 & 0 \\ 0 & 2\end{bmatrix}, \quad B=\begin{bmatrix} 1 & 2 \\ 2 & 1\end{bmatrix}.$$
We have $$A+B = \begin{bmatrix} 7 & 2 \\ 2 & 3\end{bmatrix} \ge 0, \quad A-B = \begin{bmatrix} 5 & -2 \\ -2 & 1\end{bmatrix} \ge 0$$ so $A \ge B \ge -A$. On the other hand, we have $$|A| = A, \quad |B| = \sqrt{B^*B} = \sqrt{\begin{bmatrix} 5 & 4 \\ 4 & 5\end{bmatrix}} = \begin{bmatrix} 2 & 1 \\ 1 & 2\end{bmatrix}$$ and therefore $$|A|-|B| = A-|B| = \begin{bmatrix} 4 & -1 \\ -1 & 0\end{bmatrix} \not\ge 0$$ so $|A| \ge |B|$ is false.