EDIT: A question on eigenvalues of non-negative matrices. Let $A$ and $B$ be two non-negative definite matrices, and let $l_1 \geq l_2 \geq \cdots\geq l_n \geq 0$ the eigenvalues of $A$, and $d_1 \geq d_2 \geq \cdots\geq d_n \geq 0$ the eigenvalues of $B$ then if $A-B$ is non-negative defite is true that:
$$det(A)\geq det(B)$$ and
$$l_i \geq d_i \text{ for all }i$$
Where $A$ is non negative definite if $x'Ax \geq 0$ for all posible column vector $x$.
 A: This is an answer to an older version of the question:
Regarding one part of your question: Consider $A= \begin{pmatrix} 1 & 1\\ 1&1\end{pmatrix}$ and $B=\begin{pmatrix} 1 & 0\\ 0&1\end{pmatrix}$. These are satisfying your conditions but the smallest eigenvalue of $A$ is $0$ while all eigenvalues of $B$ are $1$.
Edit: $\det(A)=0$ and $\det(B)=1$, so this is also a counterexample for the other question.
One can easily check that both matrices are positively semi-definite, as required.
A: The field wasn't specified but I asumme $\mathbb C$.
For the determinant:
(this is optional since the eigenvalue ordering proof implies this result)
$\det(A)\geq \det(B)$  is true since $A- B \succeq \mathbf 0$ we can do a unitary diagonalization so
$D_A-B'\succeq \mathbf 0\implies D_A \succeq B'$
noting this implies $0\leq b_{i,i}'\leq d_{i,i}^{(A)}$
and
$\det\big(B\big) =\det\big(B'\big) \leq \prod_{i=1}^n b_{i,i}' \leq \det\big(D_A\big) =\det\big(A\big)$
by Hadamard Determinant Inequality and then multiplying over the above point-wise bound.
For the eigenvalues:
Proceed via induction on $n$. The $n=1$ (base) case is obvious.
Inductive Case:
collect first $n-1$ orthonormal eigenvectors of B in the following $n\times n-1$
$Q:=\bigg[\begin{array}{c|c|c|c|c} \mathbf q_1 & \mathbf q_2 &\cdots & \mathbf q_{n-1}\end{array}\bigg]$
and collect last $n-1$ orthonormal eigenvectors of A in the following
$U:=\bigg[\begin{array}{c|c|c|c|c} \mathbf u_2 &\cdots & \mathbf u_{n-1} &  \mathbf u_{n}\end{array}\bigg]$
Then
$Q^*(A-B)Q\succeq \mathbf 0$
$\implies Q^*AQ \succeq Q^*BQ$
and by induction hypothesis, the eigenvalues for LHS $(\lambda_i)$ has the conjectured ordering $\lambda_i \geq d_i$ for $i \in\big\{1,2,..,n-1\big\}$.
However the eigenvalues of $Q^*AQ$ (Cauchy) interlace those of $A$ , i.e.
$l_1\geq \lambda_1\geq l_2 \geq \lambda_2\geq .... \geq l_{n-1}\geq\lambda_{n-1}\geq l_n$
thus for  $i\in \big\{1,2,...,n-1\big\}$ we have
$l_i\geq \lambda_i \geq d_i$
Then in essentially the same argument, consider
$U^*AU\succeq U^*BU$
by induction hypothesis we have
$l_i \geq \sigma_{i-1}$
for $i\in\big\{2,3,...,n\big\}$
(where $\sigma_k$ refers to kth largest eigenvalue of $U^*BU$)
By Cauchy Interlacing we have
$d_1\geq \sigma_1\geq d_2 \geq \sigma_2\geq .... \geq d_{n-1}\geq\sigma_{n-1}\geq d_n$
$\implies l_n\geq\sigma_{n-1}\geq d_n$
which completes the proof
