In general, it appears that hardly anything interesting can be said. E.g., let $A=\{(1/5,3/5),(2/5,4/5)\}$; here, it will be convenient to think of $A$ as a set of (say) open intervals, rather than a set of pairs of endpoints of intervals.

However, note first that, without loss of generality, for each open interval in $A$, all the intervals (closed, left-open, right-open) with the same endpoints may be assumed to belong to $A$. Next, let us assume that $\int_0^1|f|<\infty$ and that $A$ is a semi-ring (see measures on semi-rings)
and $[0,1]\in A$. Then the formula $\mu(I):=\int_I f$ for $I\in A$ defines a finite signed countably-additive measure $\mu$ on $A$, which can be uniquely extended to a signed measure $\bar\mu$ on the sigma-algebra $\Sigma$ generated by $A$.

The measure $\bar\mu$ determines, and is determined by, the conditional expectation $E(f|\Sigma)$ (of $f$ given $\Sigma$), equal the Radon--Nikodym derivative $\dfrac{d\bar\mu}{d\lambda|_\Sigma}$ (with respect to the underlying Lebesgue measure $\lambda$ over $[0,1]$), and this conditional expectation is then precisely all that we can get from the knowledge of the map $A\ni I\mapsto \int_I f$. (One might note that the appearance of the Radon--Nikodym *derivative* here is in broad agreement with the comment
"Recovery of function from its integral is called differentiation" by Alexandre Eremenko.)

E.g., if $A$ consists of all intervals with endpoints in the set $\{j/n\colon j=0,\dots,n\}$, then all that we will know is, in essence, the "histogram" of the average values of $f$ over the intervals $[0,1/n],\dots,[1-1/n,1]$, and this "histogram" is the best approximation to $f$ that we can get in this case.

**Extended comment:** Dirk suggested an inverse-problem approach. One may note that such an approach will work perfectly well (and, generally, even better) within the above framework of the conditional expectation. Indeed, for a space $X$ of (say) real-valued integrable functions on $[0,1]$ we have the map $X\overset K\to\mathbb R^A$ defined by the formula $Kf:=(\int_I f)_{I\in A}$ for $f\in X$. This map can be factored as follows:
\begin{equation}
X\overset{E(\cdot|\Sigma)}\longrightarrow X_\Sigma\overset{K_\Sigma}\longrightarrow\mathbb R^A,
\end{equation}
where $X_\Sigma$ is the set of all integrable $\Sigma$-measurable functions in $X$ and $K_\Sigma$ is the restriction of $K$ to $X_\Sigma$; indeed, by the definition of the conditional expectation/Radon--Nikodym derivative, we have

$K_\Sigma E(f|\Sigma)=Kf$ for all $f\in X$. Thus, instead of $K$, one can deal with its restriction $K_\Sigma$, with the same (or greater) degree of success. In particular, if $A$ is finite, then we have to deal with the finite-dimensional space $X_\Sigma$ instead of the possibly infinite-dimensional space $X$.

This comment may be viewed as an illustration of what was said previously: that the conditional expectation $E(f|\Sigma)$ is precisely all that we can get from the knowledge of the map $A\ni I\mapsto \int_I f$.