Stronger version of Acyclic Models Theorem Let $\mathscr{C}$ be an abelian category. If $P_\bullet \in \operatorname{Ch}_{\geq 0}(\mathscr{C})$ is a bounded below complex of projectives, and $C_\bullet \in \operatorname{Ch}_{\geq 0}(\mathscr{C})$ is a bounded below exact complex, then $[P_\bullet, C_\bullet] = 0$. (Every chain map $P_\bullet \to C_\bullet$ is nullhomotopic.)
It is tempting to conjecture that 

If $\mathscr{B} \underset{G}{\overset{F}{\rightrightarrows}}
 \operatorname{Ch}_{\geq 0}(\mathscr{C})$ are functors from an
  arbitrary category $\mathscr{B}$, where $F$ lands in projective
  complexes, and $G$ lands in acyclic complexes, then $[F,G]=0$, meaning
  that any natural transformation from $F \Rightarrow G$ is naturally
  chain homotopic to the zero natural transformation.

The acyclic models theorem implies something similar: that if $\mathscr{B}$ has models $\mathcal{M}$, and $F$ is a free functor w.r.t. $\mathcal{M}$, and if $G$ is acyclic, then $[F,G]$ is indeed zero.
Is the highlighted theorem above untrue?
Given a natural transformation, I can choose a map $\tau: (FX)_\bullet \to (GX)_\bullet[1]$ for each object $X \in \mathscr{B}$. Making $\tau$ natural is the problem. If I try to define the first map $\tau_0$ in the natural transformation $\tau$, and check whether it is natural, I find that the naturality diagram
\begin{array}{}
FX_0 & \xrightarrow{(Ff)_0} &FY_0\\ 
(\tau_X)_0 \downarrow && \downarrow (\tau_Y)_0
\\
GX_1 & \xrightarrow{(Gf)_1} & GY_1
\end{array}
only commutes up to a boundary element in $\partial^{GY}(GY_2)$.
 A: The trouble is that an object of the functor category $[\cal B,\cal C]$ need not be projective even if every object in the image is.  I think what you would get would be a weak homotopy (homotopy at each object of $\cal B$).  For more on this subject, we my book titled Acyclic Models, which deals with every version of the theorem I am aware of.
A: I also came up with this question recently and made some try. 
As far as I am concerned, the correct generalization of "free with basis in $\mathcal{M}$" is "projective with basis in $\mathcal{M}$".

Definition:
  A functor $S: \mathcal{C}\longrightarrow Mod_{R}$ is said to be projective with basis in $\mathcal{M}$ if the following two conditions hold



*

*$T(C)$ is projective for all $C\in\mathcal{C}$.

*There is a $T$-model set $\chi=\{x_\lambda\in T(M_\lambda)\mid M_\lambda\in \Lambda\}$ s.t.
$$
\{T(g)(x_\lambda)|g\in \hom(M_\lambda,C), \lambda\in \Lambda\}
$$
is projective basis for $T(C)$. i.e.
For each $x\in T(C)$, it can be expressed as
$$
x=\sum_{\lambda\in \Lambda}\sum_{g\in \hom(M_\lambda,C)} f_{g,\lambda}^C(x)T(g)(x_\lambda)
$$
where $\{f_{q,\lambda}^C: T(C)\longrightarrow R\}$ is a fixed set of morphisms of $R$-modules.


A functor $S_\bullet:\mathcal{C}\longrightarrow Comp_R$, where $Comp_R$ is the category of chain complex of $R$-modules, is said to be projective with basis in $\mathcal{M}$ if each $S_n$ is projective with basis in $\mathcal{M}$. 
And we can state a proposition:

Proposition:
  Suppose $\mathcal{C}$ is a category with models $\mathcal{M}$. Suppose $T_\bullet, S_\bullet:\mathcal{C}\longrightarrow Comp_R$ are two functors such that both $T_\bullet$ and $S_\bullet$ are non-negative. Assume further $T_\bullet$ is projective with basis in $\mathcal{M}$ and $S_\bullet$ is acyclic in the positive degree on each element $M\in\mathcal{M}$.
  Suppose 
  $$
\Theta: H_0\circ T_\bullet\longrightarrow H_0\circ S_\bullet
$$
   is a natural transformation. $\exists $ a  natural chain morphism $\Psi_\bullet:T_\bullet\longrightarrow S_\bullet$ which is unique up to natural chain homotopy and has $H_0(\Psi_\bullet)=\Theta$. 

And this proposition seems to be a specialization of Theorem 1 in 
Dold A., MacLane S., Oberst U. (1967) Projective classes and acyclic models. In: Reports of the Midwest Category Seminar. Lecture Notes in Mathematics, vol 47. Springer, Berlin, Heidelberg
Hope that helps.
