The answer is no, Haag's theorem does not prevent time-evolution in the Heisenberg picture.
Given any representation and any unitary transformation, applying the latter to the former gives a unitarily equivalent representation. And well-defined unitary representations do exist, as long as we're using a well-defined formulation to begin with. In particular, given a well-defined formulation with a Hilbert space, an algebra of local operators, and a Hamiltonian $H$, we can use the unitary time-evolution operator $U(t)=\exp(-iHt)$ with no problems. Haag's theorem says that if we start with the vacuum representation in a free scalar model, then no unitary transformation can give the vacuum representation of an interacting scalar model — at least not when working in infinite volume, etc; some caveats are highlighted below. So the interaction picture doesn't work, at least not in terms of operators acting on state-vectors, again with some caveats highlighted below.
What does Haag's theorem say?
For reference, this is how Haag's theorem is expressed on page 12 in "Haag's theorem in renormalised quantum field theories" (https://arxiv.org/abs/1602.00662):
Haag’s Theorem. If a scalar quantum field is unitarily equivalent to a free scalar quantum field, then, by virtue of the reconstruction theorem, it is also a free field because all vacuum expectation values coincide.
This is how it is expressed more carefully on page 49 in the same paper:
Theorem 11.7 (Haag’s Theorem). Let $\varphi$ and $\varphi_0$ be two Hermitian scalar fields of mass $m\geq 0$ in the sense of the Wightman framework. Suppose the sharp-time limits $\varphi(t,f)$ and $\varphi_0(t,f)$ exist and that at time $t = 0$ these two sharp-time fields form an irreducible set in their respective Hilbert spaces ${\cal H}$ and ${\cal H}_0$. Furthermore let there be an isomorphism $V:{\cal H}_0\rightarrow {\cal H}$ such that at time $t$, $\varphi(t,f)=V\varphi_0(t,f)V^{-1}$. Then $\varphi$ is also a free field of mass $m\geq 0$.
(I assume that $\varphi_0$ denotes a free field.) Notice that these statements of Haag's theorem are specific to scalar fields. As far as I know, the theorem has never been generalized to models that have gauge fields.
Neither of the excerpts shown above says anything that conflicts with time-evolution in the Heisenberg picture. A free scalar field remains free (with the same mass) under time-evolution, and an interacting scalar field remains interacting (with the same mass and coupling constants) under time-evolution. So Haag's theorem does not forbid time-evolution.
However, according to these excerpts, Haag's theorem does imply that a scalar field cannot begin free and then become interacting (or vice versa), so it implies that the interaction picture doesn't work — at least, it doesn't work under the theorem's strict conditions (which I didn't copy here), including strict Poincare symmetry.
Theorem 17.1 in the same paper highlights a related result that the author calls "Haag's theorem for free fields". The theorem says that two models of free scalar fields with different masses cannot be unitarily equivalent to each other.
By the way, Haag's theorem can be considered irrelevant in practice, for two reasons:
The only known mathematically well-defined constructions of most${}^{[1]}$ interacting QFTs involve treating space (or spacetime) as a finite lattice, but Haag's theorem relies on Poincare symmetry, or at least on an infinite-volume limit.
Well-defined lattice-based constructions don't use the interaction picture anyway.
We don't normally use a lattice formulation explicitly (because it's messy), but it's important in the modern view of renormalization. We can think of the usual ill-defined perturbative calculations as a convenient abbreviation for messy-but-well-defined calculations on a finite lattice. With this perspective, Haag's theorem essentially becomes irrelevant. Similar sentiments were expressed in another post about Haag's theorem.
${}^{[1]}$ In a comment, Abdelmalek Abdesselam pointed out that there are exceptions: "There are models (in 2d and 3d) constructed in the continuum and satisfying Poincare invariance."
