This is true; though I don't know a canonical reference (perhaps someone else can provide this).
But let me sketch a proof. In infinite dimensions, we need to be a little careful about topologies. Let $\newcommand{\mc}{\mathcal}\mc T(H)$ be the trace-class operators on $H$ with norm $\newcommand{\tr}{\operatorname{tr}}$$\|x\|_1 = \tr(|x|)$, and $\mc B(H)$ the algebra of all bounded operators with norm $\|\cdot\|$. As usual, $\mc T(H)$ is the predual of $\mc B(H)$ with duality $\langle x, t \rangle = \tr(xt)$.
Let $c:\mc T(H) \rightarrow \mc T(H)$ be a linear map which is completely positive. But what exactly do we mean by "completely positive" here? For $M_n(\mc T(H))$ is not a C$^*$-algebra, and so doesn't immediately have a notion of positivity (but see note 2 below). For now, we proceed as follows.
As $c$ is positive, it is bounded (see note 1). Thus we can form $c^*$, a map on $\mc B(H)$, and demand that this is completely bounded.
We now apply the usual Stinespring construction to obtain an auxiliary Hilbert space $H'$, a unital $*$-homomorphism $\pi:\mc B(H)\rightarrow\mc B(H')$, and a bounded linear map $V:H\rightarrow H'$ with $c^*(x) = V^*\pi(x)V$. We now make some observations:
- You define $\mc D(H)$ as $\{x\in\mc T(H) : x\geq 0, \tr(x)\leq 1\}$ which is just the closed unit ball of $\mc T(H)$ intersected with $\mc T(H)_+$. Notice now that the requirement that $\tr(c(x)) \leq \tr(x)$ for all $x\in \mc D(H)$ can be multiplied by any positive number, and so is equivalent to $\tr(c(x)) \leq \tr(x)$ for all $x\in\mc T(H)_+$. This is equivalent to $\langle c^*(1), x \rangle \leq \langle 1, x \rangle$ for all $x\in\mc T(H)_+$. This is equivalent to just that $c^*(1) \leq 1$ in $\mc B(H)_+$.
- Thus $c^*(1) = V^*\pi(1)V = V^*V \leq 1$.
- Careful examination of the Stinespring construction shows that $\pi$ is normal because $c^*$ is.
- A normal unital $*$-homomorphism $\mc B(H)\rightarrow\mc B(H')$ has a special form. $H' = H \otimes K$ for some Hilbert space $K$ and $\pi(x) = x\otimes 1$.
- In fact, in this case, we can streamline the Stinespring construction (see for example page 9 of arXiv:1107.5244 math.OA).
So we have $V:H\rightarrow H\otimes K$ with $V^*V\leq 1$ and $c^*(x) = V^*(x\otimes 1)V$. Let $(e_i)$ be an orthonormal basis of $K$, and check that there are $E_i\in\mc B(H)$ with
$$ V(\xi) = \sum_i E_i(\xi)\otimes e_i \qquad (\xi\in H). $$
That $V^*V\leq 1$ means exactly that $\sum_i E_i^*E_i\leq 1$. Then
$$ c^*(x) = \sum_i E_i^* x E_i. $$
It is now easy to check that $c(t) = \sum_i E_itE_i^*$ to $t\in\mc T(H)$.
Finally, if $H$ is separable, then $K$ will be separable and so the family $(E_i)$ will be a sequence (or finite).
Note 1: If $c$ is positive but not bounded, then firstly observe that every $t\in\mc T(H)$ can be written as the linear combination of 4 positive trace-class operators, so $c$ is not bounded on the positives. Thus we can find $(x_n)\subseteq \mc T(H)_+$ with $\tr(x_n)\leq n^{-2}$ but $\tr(c(x_n))\geq n$. Then $x=\sum_n x_n$ converges absolutely in $\mc T(H)_+$, and for any $N$, $\sum_{n\leq N}x_n \leq x$ so $\sum_{n\leq N} c(x_n) \leq c(x)$ so $\tr(c(x)) \geq \sum_{n\geq N} \tr(c(x_n)) \geq N$, a contradiction.
Note 2: We identify $M_n(\mc B(H))$ with $\mc B(H\otimes \mathbb C^n)$ which gives the canonical $C^*$-algebra structure on $M_n(\mc B(H))$ and hence the positive cone. The predual is $\mc T(H\otimes \mathbb C^n)$ which is of course just $M_n(\mc T(H))$. Furthermore, duality behaves as you expect, namely if $x=(x_{i,j})\in M_n(\mc B(H))$ and $t=(t_{i,j}) \in M_n(\mc T(H))$ then
$$ \tr(xt) = \sum_{i,j} \tr(x_{i,j} t_{j,i}). $$
Then $c^*$ being completely positive is equivalent to $c$ being completely positive, where $M_n(\mc T(H))$ is given this positive cone, i.e. as operators on $H\otimes\mathbb C^n$. (Be aware that it is also common to use $\langle x,t\rangle = \sum_{i,j} \langle x_{i,j} , t_{i,j} \rangle = \sum_{i,j} \tr(x_{i,j}t_{i,j})$ which is of course different.)
