No - though you are right in thinking that a counterexample is a bit hard to construct: in particular, something like a step function such as
$$f_1 : \mathbb{R} \rightarrow \mathbb{R},\ f_1(x) := \begin{cases} 0\ \mbox{if $x < 0$}, \\1\ \mbox{if $x \ge 0$}\end{cases}$$
is not antidifferentiable on its whole domain, because at all domain points except $0$, any antiderivative of such a function has to satisfy
$${F_1}_\mathrm{putative}(x) = \begin{cases} C_1\ \mbox{if $x < 0$}\\x + C_2\ \mbox{if $x > 0$}\end{cases}$$
for some constants $C_1$ and $C_2$ - but there is no way to define a value exactly at $0$ that will make this differentiate to $f_1$ there, because this function is simply not differentiable at $0$ (the closest you can get is if $C_1 = C_2 = C$, and then if you define it such that ${F_1}_\mathrm{putative}(0) = C$, it will be continuous; but it is not differentiable there as it has a corner there).
Instead, you have to get a bit cleverer. A standard counterexample is (if I remember right)
$$f_2: \mathbb{R} \rightarrow \mathbb{R}, f_2(x) := \begin{cases} \sin\left(\frac{1}{x}\right)\ \mbox{if $x \ne 0$}\\ 0\ \mbox{if $x = 0$} \end{cases}$$
The antiderivatives of this function satisfy
$$F_2(x) = \begin{cases}x \sin\left(\frac{1}{x}\right) - \mathrm{Ci}\left(\frac{1}{x}\right) + C\ \mbox{if $x \ne 0$}\\
C\ \mbox{if $x = 0$}\end{cases}$$
including at $x = 0$, for some constant $C$. It can be checked that these functions are differentiable at $x = 0$, with derivative $F'_2(0) = 0$.
ADD: Looks like I remembered wrong - commenters above showed as more "standard" example that where we start with $F_3(x) := \begin{cases}x^2 \sin\left(\frac{1}{x}\right)\ \mbox{if $x \ne 0$}\\ 0\ \mbox{if $x = 0$}\end{cases}$ as antiderivative ... but meh! The above works and its exposition to me follows the direction of the implication in the question better: construct a weird "derivative" and then try and antidifferentiate it.
