I would say that it is endothermic, and then by increasing the temperature, one could get more product, disturbing the equilibrium.
Is this true, though? During an esterification, some bonds are broken, releasing energy. The opposite reaction splits the ester and adds the atoms back, which I suppose releases energy which is then bound in the bonds adding back the water molecule.
When in doubt, go find data. Here is a typical such reaction - the esterification of ethanol and acetic acid to make ethyl acetate.
$$\ce{CH3CH2OH + CH3CO2H -> CH3CO2CH2CH3 + H2O}$$
The following thermodynamic values are from the NIST Chemistry Webbook.
$$\begin{array}{|c|c|}\hline \text{compound} & \Delta _\pu{f} H^\circ \ (\pu{kJ mol^-1}) \\ \hline \text{ethanol} & -276 \\ \text{acetic acid} & -484 \\ \text{ethyl acetate} & -480 \\ \text{water} & -286 \\ \hline \end{array}$$
Now for some Hess's Law:
$$\begin{aligned} \Delta_\pu{r} H^\circ &= \sum_\mathrm{products} \Delta_\pu{f} H^\circ - \sum_\mathrm{reactants} \Delta_\pu{f} H^\circ \\ &= (-480\ \pu{kJ mol^-1} + -286\ \pu{kJ mol^-1}) - (-276\ \pu{kJ mol^-1} + -484\ \pu{kJ mol^-1}) \\ &= -766\ \pu{kJ mol^-1} - (-760\ \pu{kJ mol^-1}) \\ &= -6\ \pu{kJ mol^-1}\end{aligned}$$
Hey! This one is exothermic (slightly).
Here are the enthalpies of reaction for some other esters (by no means a complete or even representative list): $$\begin{array}{|c|c|}\hline \mathrm{ester} & \Delta_\pu{r} H^\circ \ (\pu{kJ mol^-1}) \\ \hline \text{methyl acetate} & -9 \\ \text{ethyl acetate} & -6\\ \text{methyl benzoate} & -5 \\ \text{ethyl crotonate} & +1 \\ \hline \end{array}$$
I would thus guess that formation of most carboxylic esters from carboxylic acids and alcohols would be slightly exothermic or slightly endothermic depending on structure.
The reasons esterifications are often heated are 1) to increase the rate of reaction and 2) to distill off water as it forms.
