Balanced categories and conservative functors. If $\mathbf{C}$ is a balanced category, then every faithful functor out of $\mathbf{C}$ is conservative.
Proof. Suppose $\mathbf{C}$ is a balanced category and let $F$ denote a faithful functor out of $\mathbf{C}.$ Suppose $\varphi$ is an arrow of $\mathbf{C}$ such that $F(\varphi)$ is an isomorphism. Then $F(\varphi)$ is both monic and epic. By faithfulness, this means that $\varphi$ is both monic and epic. By balancedness, this means that $\varphi$ is an isomorphism. Hence $F$ is conservative.

Question. Does the converse hold?
I.e. if every faithful functor out of $\mathbf{C}$ is conservative, is $\mathbf{C}$ necessarily balanced?

 A: Following Qiaochu's suggestion, we can obtain a result of the kind you are looking for.
Say a morphism in $\mathcal{C}$ is a weak equivalence if it is both monic and epic. Then:

The following are equivalent:
  
  
*
  
*$\mathcal{C}$ is a balanced category.
  
*Every faithful functor $\mathcal{C} \to \mathcal{D}$ is conservative, and the weak equivalences in $\mathcal{C}$ satisfy the right Ore condition, i.e. for every morphism $f : X \to Y$ in $\mathcal{C}$ and every weak equivalence $y : Y' \to Y$ in $\mathcal{C}$, there is a commutative square of the form below,
  $$\require{AMScd}
\begin{CD}
X' @>{x}>> X \\
@V{f'}VV @VV{f}V \\
Y' @>>{y}> Y
\end{CD}$$
  where $x : X' \to X$ is also a weak equivalence in $\mathcal{C}$.
  

The downward implication is straightforward, so assume the second set of hypotheses. Since weak equivalences in $\mathcal{C}$ are monomorphisms, the right Ore condition suffices to establish a calculus of right fractions. Concretely, that means we have the following construction of the localisation of $\mathcal{C}$ with respect to weak equivalences:


*

*The objects are as in $\mathcal{C}$.

*The morphisms $X \to Y$ are equivalence classes zigzags of the form below,
$$\begin{CD}
X @<<< \bullet @>>> Y
\end{CD}$$
where the leftward-pointing arrow is a weak equivalence in $\mathcal{C}$, and given any commutative diagram in $\mathcal{C}$ of the form below,
$$\begin{CD}
X @<<< \bullet @>>> Y \\
@| @AAA @| \\
X @<<< \bullet @>>> Y \\
@| @VVV @| \\
X @<<< \bullet @>>> Y
\end{CD}$$
where all the leftward-pointing arrows are weak equivalences in $\mathcal{C}$, we define the top row and bottom row to be equivalent zigzags.


Call this category $\mathcal{D}$. There is an evident functor $\mathcal{C} \to \mathcal{D}$ and it is faithful: indeed, given a commutative diagram of the form below,
$$\begin{CD}
X @= X @>>> Y \\
@| @AAA @| \\
X @<<< X' @>>> Y \\
@| @VVV @| \\
X @= X @>>> Y
\end{CD}$$
where $X' \to X$ is an epimorphism, the two arrows $X \to Y$ must be equal. 
Thus, under the hypothesis that every faithful functor $\mathcal{C} \to \mathcal{D}$ is conservative, every weak equivalence in $\mathcal{C}$ is already an isomorphism, i.e. $\mathcal{C}$ is balanced.
It may be worth noting that there is a non-balanced category $\mathcal{C}$ for which the above right Ore condition is satisfied, namely $\mathbf{Top}$. But there is an obvious faithful non-conservative functor $\mathbf{Top} \to \mathbf{Set}$, so we haven't learned much.
A: If this is true, the following seems to me like the only proof strategy that could work. Suppose $f : c \to d$ is monic and epic. Then you should try to show that the localization functor $C \to C[f^{-1}]$ is faithful (presumably using the fact that $f$ is both left and right cancellative). If that's true, then by hypothesis it's conservative, so since $f$ is an iso in this localization, it was already an iso in $C$. 
Unfortunately I'm not sure how to show this off the top of my head. 
