Let $A$ and $B$ be non-negative ($(A x, x) \geq 0$ for all $x \in \mathcal{D}(A)$, similarly for $B$) densely defined self-adjoint operators on a Hilbert space $H$. Then the spectral theorem defines fractional powers $A^r$ and $B^r$ as non-negative self-adjoint operators on $H$, where we assume $0 < r < 1$. My question is, if we have $\mathcal{D}(A) \subset \mathcal{D}(B)$, do we necessarily have $\mathcal{D}(A^r) \subset \mathcal{D}(B^r)$, or is additional data needed?

Yes, this works. By the closed graph theorem, applied to $\textrm{id}:D(A)\to D(B)$ endowed with the operator norms, your assumption that $D(A)\subseteq D(B)$ implies that $B$ is $A$-bounded. So $$ \|Bx\| \le C_0 (\|Ax\| + \|x\|) , $$ and this implies that $B^2\le C^2(A+1)^2$, as positive definite operators (to confirm this, just write out this condition and use that $\|Bx\|^2=\langle x, B^2 x\rangle$ etc.). Now $t\mapsto t^{r/2}$ is an operator monotone function, so we also have that $B^r\le C^r (A+1)^r$ and since $D(A^r)=D((A+1)^r)$, this gives the claim.