Here's something we can say which addresses a large class of examples.

**Claim:** Let $R$ be a commutative ring which is not zero-dimensional (i.e. there exists $r \in R$ which is not a unit or a zero-divisor). Then the derived category $ho(D(R))$ of left $R$-modules is not concrete (i.e. does not admit a faithful functor to $Set$) and in particular is not accessible.

The proof is a straightforward generalization of Freyd's argument that the homotopy category of spaces is not concrete, and in particular is robust enough that "$D(R)$" could be interpreted to have various "boundedness" conditions if one likes.

**Beginning of Proof of Claim:** Pick $r \in R$ which is neither a unit nor a zero-divisor. If $M$ is a left $R$-module, define the submodule $r^\alpha M$ for any ordinal $\alpha$ inductively by $r^0 M = M$, $r^{\alpha+1} M = rr^\alpha M$, and taking intersections at limit ordinals. Note that if $\phi: M \to N$ is any $R$-module map, then $\phi(r^\alpha M) \subseteq r^\alpha N$ for any $\alpha$.

Let $\alpha$ be an ordinal. For each $n \in \mathbb N$, let $W_\alpha^{(n)}$ be the set of strictly increasing sequences of ordinals $\alpha_0 < \alpha_1 < \dots < \alpha_m$ where $m \leq n$ and $\alpha_m \leq \alpha$. Let $F_\alpha^{(n)} = R\{W_\alpha^{(n)}\}$ be the free left $R$-module on generators given by $W_\alpha^{(n)}$, and define $M_\alpha^{(n)} = F_\alpha^{(n)} / K_\alpha^{(n)}$, where $K_\alpha^{(n)}$ is generated by those elements of the form $[\alpha_1,\dots,\alpha_n] - r[\alpha_0, \alpha_1, \dots, \alpha_n]$ (when $n = 0$ we interpret this to mean that $r[\alpha_0] \in K_\alpha^{(n)}$).

**Lemma:** For $n \in \mathbb N$, the map $M_\alpha^{(n)} \to M_\alpha^{(n+1)}$ is injective.

**Proof:** This is straightforward, using the fact that $r$ is not a right zero-divisor.

Define $M_\alpha = \cup_{n \in \mathbb N} M_\alpha^{(n)}$. We have a natural filtration $M_\alpha^{(0)} \subseteq \cdots \subseteq M_\alpha^{(n)} \subseteq \cdots M_\alpha$, and the associated graded is naturally identified with $\oplus_{n \in \mathbb N} \oplus^{ W_\alpha^{(n)} \setminus W_\alpha^{(n-1)}} (R/(Rr))$.

**Lemma:** Pick a set of "canonical" coset representatives in $R$ of the nonzero elements of $R/(Rr)$. Then every element of $M_\alpha$ may be written uniquely in the form $\sum_i r_i w_i$ where the $r_i$ are canonical coset representatives and the $w_i$ are distinct words of some $W_\alpha^{(n)}$.

**Proof:** The existence of such a representation is basically obvious. Suppose that two such representations denote the same element of $M_\alpha^{(n)}$, where $n$ is minimal: $\sum_i r_i w_i = \sum_j s_j v_j$. Then they have the same image in the associated graded, and so the $w_i$'s of length $n$ match up with the $v_j$'s of length $n$; since the choice of coset representatives has been normalized, their coefficients in fact coincide. Deleting these words, we get an identification of represntations in $M_\alpha^{(n-1)}$, contradicting the minimality of $n$.

**Lemma:** For all ordinals $\beta$, $r^\beta M_\alpha$ consists of those elements whose canonical representation as chosen above contains only words $[\alpha_0,\dots,\alpha_n]$ where $\alpha_0 \geq \beta$.

**Proof:** Using the canonical representation, this is now an easy transfinite induction. (Here we use the fact that $R$ is commutative, though.)

**Conclusion of Proof of Claim:** Note that $r^\alpha M_\alpha = R/(Rr) \neq 0$ canonically, because $r$ is not a right unit. Let $f_\alpha$ be the map $R/(Rr) = r^\alpha M_\alpha \subseteq M_\alpha$. Then for $\alpha < \beta$, there is no factorization of $f_\beta$ through $f_\alpha$. Pick $d \in \mathbb Z$ such that $(R/r)[d]$, each $M_\alpha[d]$, and the fiber $(M_\alpha/f_\alpha)[d-1]$ of $f_\alpha[d]$ are all in whichever flavor of $ho(D(R))$ we are working with. Then the maps $f_\alpha[d]$ are a proper class of pairwise-inequivalent weak cokernels of their fibers in $ho(D(R))$. But as Freyd shows, a proper class of pairwise-inequivalent weak cokernels out of a fixed object in a pointed category imply that the category is not concrete.

So the outline, as in Freyd's proof, is to use the fact that in $ho(D(R))$, basically every map is a weak cokernel (in fact, all maps are if we're working in the unbounded derived category).