Components of Stress-Energy Tensor, in any arbitary coordinates, are given by, $$ T_{{\mu}{\nu}} = T(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}) .$$
One can physically interpret them as follows: $ T_{{\mu}{\nu}}$, at a point P of space-time, tells the flow of $\mu ^{th} $ component of four momentum along the $\frac{\partial}{\partial x^{\nu}}$ direction. For example, $T_{00}$ denotes how much energy per unit volume is flowing in time direction, which is same as energy density. Similarly $T_{i0}$ denotes flow of momentum (not four momentum) per unit volume along time direction, that is momentum density.
Thus, $T_{ii}$ denotes flow of $i^{th}$ component of momentum along $\frac{\partial}{\partial x^i}$ direction. But that's the definition of pressure. Since pressure is a local phenomenon, even in curved space-time, it does not matter whether you work in curvilinear or rectilinear coordinates. Locally every transformation is linear enough to define pressure as we usually do.
In your example, $(\frac{\partial}{\partial r},\frac{\partial}{\partial \theta},\frac{\partial}{\partial \phi})$, at a point could be thought of constituting a Cartesian system. The radial direction could very well be defined as x direction, locally. Since pressure is also local, $T_{rr}$ denotes pressure along the radial direction. Same goes with other two directions. $T_{\theta \theta}$ and $T_{\phi \phi}$ denotes pressure along $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \phi} $ directions respectively.
For the second part of your question, No, you cannot do that. That would break spherical symmetry. A concise explanation is: metric and matter are coupled by Einstein Field Equation. If one is not symmetric under rotations then other cannot be symmetric.
A detailed explanation: Assume metric is spherically symmetric and matter is given by, $$T_{\mu\nu}=diag(\,\rho(r)\; e^{\nu(r)},p_{r}(r)\;e^{\lambda(r)}, p_{\theta}(r)\;r^{2},p_{\theta}(r)\;r^{2}\sin^{2}\theta\,).$$ If metric is invariant under rotation so is Einstein Tensor, $G_{\mu \nu}$, as Einstein Tensor is made from metric tensor. We have following equation, $$ G_{\mu \nu}=\kappa T_{\mu \nu} .$$ The left hand side is invariant under rotation but right hand side is not. A contradiction! Therefore our assumption is wrong. Matter cannot be given by our proposed form.
Best Answer
We adopt the system of units in which speed of light is 1.
Components of the stress tensor $T^{\alpha\beta}$ physically mean the following: $T^{00}$ is the energy density, $T^{0j}$ is the energy flux across the spatial-surface $x_j=$ constant ($j=1,2,3$), $T^{i0}$ is the density of $i$-th component of momentum, and $T^{ij}$ is the $i$-th component of momentum flux across the spatial-surface $x_j=$ constant ($i,j=1,2,3$). Normal momentum flux ($T^{ij}$ for $i=j$) causes normal stress on the fluid element and the others ($T^{ij}$ for $i\neq j$) cause shear stress on the fluid element.
An ideal fluid is one whose viscosity and conductivity are zero. Consider an elemental volume of ideal fluid in its MCRF (momentarily co-moving reference frame). Since conductivity is zero, there is no energy flux into or out of it, which implies $T^{0j}=0$. Since there is no viscosity, it doesn't experience shear stresses, therefore $T^{ij}=0$ when $i\neq j$. Further the statement that the fluid has no viscosity is a frame-independent statement, so $T^{ij}=0$ when $i\neq j$ in any reference frame, and so the matrix $T^{ij}$ must be diagonal in all reference frames. This is possible only if $T^{ij}=p\delta^{ij}$ in which $\delta^{ij}$ is the identity tensor and $p$ is a scalar called pressure. If we denote the energy density by $\rho$, then the stress tensor $T^{\alpha\beta}$ in the MCRF of the fluid element is: $$\begin{bmatrix} \rho & 0 & 0& 0\\ 0 & p& 0& 0\\ 0 & 0& p& 0\\ 0 & 0& 0& p \end{bmatrix}$$ This can be simplified as: $$\begin{bmatrix} \rho & 0 & 0& 0\\ 0 & p& 0& 0\\ 0 & 0& p& 0\\ 0 & 0& 0& p \end{bmatrix}= \begin{bmatrix} \rho+p & 0 & 0& 0\\ 0 & 0& 0& 0\\ 0 & 0& 0& 0\\ 0 & 0& 0& 0 \end{bmatrix}+ \begin{bmatrix} -p & 0 & 0& 0\\ 0 & p& 0& 0\\ 0 & 0& p& 0\\ 0 & 0& 0& p \end{bmatrix}\\ \Rightarrow\quad T^{\alpha\beta}=(\rho+p)(\mathbf{e}_0\mathbf{e}_0)^{\alpha\beta}+p\eta^{\alpha\beta}$$
in which $\eta^{\alpha\beta}$ is the metric tensor. The unit vector in the time direction (of the MCRF of the fluid element) $\mathbf{e}_0$ is nothing but its 4-velocity $\mathbf{U}$. Therefore the dyadic $\mathbf{e}_0\mathbf{e}_0=\mathbf{U}\mathbf{U}$, whose component is $(\mathbf{U}\mathbf{U})^{\alpha\beta}=U^\alpha U^\beta$. Thus we have: $$T^{\alpha\beta}=(\rho+p)U^\alpha U^\beta+p\eta^{\alpha\beta}$$
Reference: General Relativity by B. Schutz.