In short:
The pressure $p$ arises from random motions of individual particles. The velocity $U^\mu$ does not capture these individual random motions. It is rather an average velocity over all particles, or drift of the fluid. In this average sense the fluid follows geodesics when pressure is constant.
If $p$ is not constant it is however no longer true that the fluid moves along geodesics. The pressure gradient leads to an additional force different from gravitation.
Zero Pressure
To see how geodesics arise when $p=0$ recall that the energy--momentum tensor of dust (or a pressure--less perfect fluid) is
\begin{eqnarray} T^{\mu\nu}=\rho\,U^\mu U^\nu\,. \end{eqnarray} The law of conservation of energy-momentum in curved spacetime is
\begin{eqnarray}\tag{1} 0=\nabla_\mu T^{\mu\nu}=\rho\,U^\nu\nabla_\mu U^\mu + \rho\,U^\mu\nabla_\mu U^\nu + (\partial_\mu \rho)\,U^\mu U^\nu\,. \end{eqnarray} Multiplying this with $U_\nu$ and summation convention yields
\begin{eqnarray} 0&=&\rho\,\underbrace{U^\nu U_\nu}{=-1}\nabla\mu U^\mu + \rho\,U^\mu\underbrace{U_\nu\nabla_\mu U^\nu}{=0}+(\partial\mu \rho)\,U^\mu \underbrace{U_\nu U^\nu}{=-1}\ &=&-\rho\,\nabla\mu U^\mu -(\partial_\mu \rho)\,U^\mu=-\nabla_\mu (\rho\, U^\mu)\,, \end{eqnarray} i.e., the four--divergence $\nabla_\mu (\rho\,U^\mu)$ is zero. 649
Combining the first and last term in the energy conservation law this can clearly be written as
\begin{eqnarray} 0= U^\nu \nabla_\mu (\rho\, U^\mu) + \rho\,U^\mu\nabla_\mu U^\nu\,. \end{eqnarray} Because we have just shown that the four-divergence in the first term is zero this collapses to
\begin{eqnarray} 0=\rho\, U^\mu\nabla_\mu U^\nu \,. \end{eqnarray} It is clear that in regions of non vanishing $U^\mu$ the density $\rho$ cannot be zero. Therefore,
\begin{eqnarray}\tag{2} 0=U^\mu\nabla_\mu U^\nu \,. \end{eqnarray} This equation says that the four-momentum vector field is the tangent vector field of timelike geodesics. In other words, the fluid particles move along geodesics.
Non Zero Pressure
When there is non zero pressure $p$ the energy--momentum tensor of the fluid is
\begin{eqnarray} T^{\mu\nu}=(\rho+p)\,U^\mu U^\nu+p\,g^{\mu\nu}\,. \end{eqnarray} If $p$ is constant then $\nabla_\mu(p\,g^{\mu\nu})$ is zero (because $\Gamma^\lambda_{\mu\nu}$ is metric compatible) and the energy conservation leads again to (2). 649