You can define a new macro that provides a slightly displaced dot, say \dispdot[<disp>]{<stuff>}. Below I've defined \dispdot to give a .2ex displacement. Supplying a different <disp> will change the location of the \dot:

\documentclass{article}
\newcommand{\dispdot}[2][.2ex]{\dot{\raisebox{0pt}[\dimexpr\height+#1][\depth]{$#2$}}}% \dispdot[<displace>]{<stuff>}
\begin{document}
$d\mathcal{U}, d\dot{\mathcal{U}}, d\dispdot{\mathcal{U}}, d\dispdot[.5ex]{\mathcal{U}}$
\end{document}
If need be, it would be possible to capture the math style.
For horizontal displacement, you could define \dispdot to give a 3mu displacement (actually, due to symmetry it is half the actual displacement):

\documentclass{article}
\newcommand{\dispdot}[2][3mu]{\dot{#2\mkern#1}\mkern-#1}% \dispdot[<disp>]{<stuff>}
\begin{document}
$d\mathcal{U}, d\dot{\mathcal{U}}, d\dispdot{\mathcal{U}}, d\dispdot[5mu]{\mathcal{U}}$
\end{document}
Note that math units are very small, so adjustments will be very tiny.
palatinopackage doesn't show that symbol, as it doesn't affect math. Please, add a minimal example showing how you get that output. You should try also\usepackage{newpxtext,newpxmath}or\usepackage{mathpazo}for using math with the Palatino font and not\usepackage{palatino}. – egreg Aug 15 '13 at 09:15