Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are:
For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\bullet)$ are derived $I$-complete.
For any nonzero derived $I$-complete $A$-module $B$, the $A/I$-module $B/IB$ is nonzero.
Your desired assertion follows from 1. and 2. (apply assertion 2. to the highest cohomology module $B$ of your complex $C^\bullet$, taking into account that $B$ is derived complete by assertion 1.)
Concerning the proof of 2., you can proceed by induction in the number of generators $s_1,\dotsc,s_m$ of the ideal $I$: prove that $B\ne0$ implies $B/s_1B\ne0$ implies $B/(s_1B+s_2B)\ne0$ etc. Here it helps to observe that:
If $B$ is derived $I$-complete and $J\subset I$, then $B$ is derived $J$-complete.
The kernel and cokernel of any morphism of derived $I$-complete modules is derived $I$-complete. (This is just a restatement of 1.)
The observations 3. and 4. reduce your question to proving that $B/sB\ne0$ whenever $B\ne0$ is a derived $(s)$-complete $A$-module, for the principal ideal $(s)$ generated by a single element $s\in A$.
At this point you recall that $B$ being derived $(s)$-complete means that $Hom_A(A[s^{-1}],B)=0=Ext^1_A(A[s^{-1}],B)$. It remains to check that the equation $B=sB$ implies surjectivity of the natural map $Hom_A(A[s^{-1}],B)\to B$, which is a straightforward exercise. So $Hom_A(A[s^{-1}],B)=0$ together with $sB=B$ imply $B=0$.
References:
Bhatt's short paper https://doi.org/10.1016/j.jpaa.2018.08.008 .
My long paper https://arxiv.org/abs/1605.03934 (in which "derived complete modules" are called "contramodules").
EDIT: A question was asked in the comments whether the assertion remains true for unbounded complexes. The answer is positive, but the proof below is more complicated.
Let $s_1,\dotsc,s_m$ be a finite set of generators of the ideal $I\subset A$. Denote the sequence $s_1,\dotsc,s_m$ by the single letter $\mathbf s$ for brevity. The augmented Cech complex $\check C(A;\mathbf s)$ is defined as the tensor product
$$
\check C(A;\mathbf s)=(A\to A[s_1^{-1}])\otimes_A\dotsb
\otimes_A(A\to A[s_m^{-1}]).
$$
Consider the unbounded derived category of $A$-modules $D(A{-}Mod)$, and denote by $D_{I-tors}(A{-}Mod)\subset D(A{-}Mod)$ the full triangulated subcategory of complexes of $A$-modules with $I$-torsion cohomology modules. Denote by $D_{I-com}(A{-}Mod)\subset D(A{-}Mod)$ the full triangulated subcategory of complexes of $A$-modules with derived $I$-adically complete cohomology modules. The full subcategory $D_{I-com}(A{-}Mod)$ coincides with what is usually called the full subcategory of derived $I$-adically complete complexes in $D(A{-}Mod)$.
The key fact for the argument below is that there is an equivalence of categories
$$
D_{I-tors}(A{-}Mod)\simeq D_{I-com}(A{-}Mod)
$$
provided by the functor of tensor product with the augmented Cech complex
$$
\check C(A;\mathbf s)\otimes_A{-}\colon D_{I-com}(A{-}Mod)
\longrightarrow D_{I-tors}(A{-}Mod),
$$
whose quasi-inverse is the derived Hom functor
$$
\mathbf R Hom_A(\check C(A;\mathbf s),{-})\colon D_{I-tors}(A{-}Mod)
\longrightarrow D_{I-com}(A{-}Mod).
$$
This result is known as the Matlis-Greenlees-May (MGM) duality or Matlis-Greenlees-May equivalence.
Specifically, it is important for us that, for any complex of $A$-modules $B^\bullet$ with derived $I$-adically complete cohomology modules (or in other words, for any derived $I$-adically complete complex $B^\bullet$), the complex $\check C(A;\mathbf s)\otimes_A B^\bullet$ is not acyclic if the complex $B^\bullet$ is not acyclic.
Let us prove that the complex $B^\bullet$ is acyclic if the complex $A/I\otimes_A^{\mathbf L}B^\bullet$ is. First of all, we can replace $B^\bullet$ by its homotopy flat ("K-flat") resolution. This allows us to assume that $B^\bullet$ is a homotopy flat complex (to simplify matters, one can additionally assume $B^\bullet$ to be a complex of flat $A$-modules).
Then we have $A/I\otimes_A^{\mathbb L} B^\bullet=A/I\otimes_A B^\bullet$, that is the derived tensor product coincides with the underived one. Assume that the complex $A/I\otimes_A B^\bullet$ is acyclic.
Since $B^\bullet$ is a homotopy flat complex of $A$-modules, $A/I\otimes_A B^\bullet$ is also a homotopy flat complex of $A/I$-modules. It follows that the complex $N\otimes_A B^\bullet = N\otimes_{A/I}(A/I\otimes_A B^\bullet)$ is acyclic for any $A/I$-module $N$. Hence the complex $N\otimes_A B^\bullet$ is also acyclic for any $A/I^k$-module $N$, where $k\ge1$ is any integer. Passing to the direct limit, we can conclude that the complex $N\otimes_A B^\bullet$ is acyclic for any $I$-torsion $A$-module $N$.
Finally, it follows that the complex $M^\bullet\otimes_A B^\bullet$ is acyclic for any finite complex of $A$-modules $M^\bullet$ with $I$-torsion cohomology modules. For example, $M^\bullet = \check C(A;\mathbf s)$ is such a complex. We have shown that the complex $\check C(A;\mathbf s)\otimes_A B^\bullet$ is acyclic. According to the above MGM equivalence theorem, it follows that the complex $B^\bullet$ is acyclic.
References to MGM duality/equivalence:
Dwyer, Greenlees "Complete modules and torsion modules", https://doi.org/10.1353/ajm.2002.0001
My paper "Dedualizing complexes and MGM duality", https://arxiv.org/abs/1503.05523 , https://doi.org/10.1016/j.jpaa.2016.05.019 , Section 3.
![taken from the paper [PSY]](https://i.sstatic.net/obvfp.png)
