It is very common for a topological space to be a complete uniform space in some uniformity, but it is less common for a topological space to satisfy the Baire category theorem since the proof of the Baire category theorem makes an essential use of countability of the metric rather than the uniformity..
Every paracompact space has a compatible complete uniformity simply by taking the uniform covers to be open covers and the entourages to be the neighborhoods of the diagonal. In fact, the paracompact spaces are precisely the spaces that have a compatible uniformity that satisfies a stronger form of completeness. A uniform space is said to be supercomplete if its hyperspace is complete. A topological space has a compatible supercomplete uniformity if and only if it is paracompact.
A space whose cardinality is below the first measurable cardinal is completely uniformizable if and only if the space is realcompact.
Every countable completely regular space without any isolated points $X$ can be given a compatible complete uniformity, but $X$ is never a Baire space. For example, $\mathbb{Q}$ can be given a compatible complete uniformity, but $\mathbb{Q}$ is the countable union of nowhere dense subspaces.
Every good general topology text will tell you that an $F_{\sigma}$-set in a paracompact space is paracompact. The spaces mentioned by Priel in his answer are paracompact since they are countable unions of closed subspaces of a paracompact space (the space of all smooth functions with compact support is metrizable since its topology is induced by countably many seminorms, namely $Sup_{x\in\mathbb{R}} |f^{(n)}(x)|$ and every metrizable space is paracompact).
If $X$ is a completely regular space, then let $Ba(X)$ denote the $\sigma$-algebra of Baire sets. In other words, $Ba(X)$ is the $\sigma$-algebra on $X$ generated by the zero sets (Recall that a zero-set is a set of the form $f^{-1}[\{0\}]$ for some continuous $f:X\rightarrow\mathbb{R}$ and that the complement of a zero set is a cozero set).
$\mathbf{Lemma}$ A completely regular space $X$ is realcompact if and only if each $\sigma$-complete ultrafilter on $Ba(X)$ is of the form $\{R\in Ba(X)|x_{0}\in R\}$ for some
$\mathbf{Proposition}$ Every $F_{\sigma}$-set in a realcompact space is also realcompact.
$\mathbf{Proof}:$ Suppose that $X$ is a realcompact space and $C_{n}\subseteq X$ is a closed subspace for each natural number $n$. Let $D=\bigcup_{n\in\omega}C_{n}$. Let $\mathcal{M}\subseteq Ba(D)$ be a $\sigma$-complete ultrafilter. Then define $\mathcal{N}=\{R\in Ba(X)|R\cap D\in\mathcal{M}\}$. Then $\mathcal{N}$ is a $\sigma$-complete ultrafilter on $Ba(X)$. Therefore, $\mathcal{N}=\{R\in\mathcal{M}|x_{0}\in R\}$ for some $x_{0}\in X$. I claim that $x_{0}\in D$.
Suppose to the contrary that $x_{0}\not\in D$. Then since $x_{0}\in\overline{D}$ and $X$ is completely regular, for all $n$ there is some cozero set $U_{n}\subseteq X$ with $x_{0}\in U_{n}$ but $U_{n}\cap C_{n}=\emptyset$. Therefore, $U_{n}\in\mathcal{N}$ for each $n$, so $\bigcap_{n}U_{n}\in\mathcal{N}$ as well. However, since $\bigcap_{n}U_{n}\in\mathcal{N}$, we have $\emptyset=D\cap\bigcap_{n}U_{n}\in\mathcal{M}$ which is a contradiction. Therefore, $x_{0}\in D$.
I now claim that $\mathcal{M}=\{R\in Ba(D)|x_{0}\in R\}$. Suppose that $R\in Ba(D)$ and $x_{0}\in R$. Then there are open subsets $V_{n}\subseteq D$ such that $x_{0}\in V_{n}$ for all $n$ but where $\bigcap_{n} V_{n}\subseteq R$. Therefore, for all $n$, there is some open subset $V_{n}^{\sharp}\subseteq X$ such that $V_{n}=V_{n}^{\sharp}\cap D$. Therefore, since $x_{0}\in V_{n}^{\sharp}$, for all $n$, there is a cozero set $W_{n}^{\sharp}\subseteq X$ such that $x_{0}\in W_{n}^{\sharp}\subseteq V_{n}^{\sharp}$. Let $W_{n}=W_{n}^{\sharp}\cap D$. Then $W_{n}^{\sharp}\in\mathcal{M}$ for all $n$, so $W_{n}=W_{n}^{\sharp}\cap D\in\mathcal{N}$. Therefore $\bigcap_{n\in\omega}W_{n}\in\mathcal{N}$. However, since $W_{n}\subseteq V_{n}$ and $\bigcap_{n\in\omega}W_{n}\subseteq\bigcap_{n}V_{n}\subseteq R$, we have $R\in\mathcal{N}$ as well. We therefore conclude $\mathcal{N}=\{R\in Ba(D)|x_{0}\in R\}$. Therefore $D$ is realcompact. $\mathbf{QED}$
Using the above proposition, one can obtain a realcompact space that does not satisfy the Baire category theorem from any completely regular space that does not satisfy the Baire category theorem: Suppose that $X$ is a completely regular space which is the union of countably many nowhere dense sets $(L_{n})_{n\in\omega}$. Let $Y$ be a realcompact space such that $X$ is a dense subspace of $Y$ (for example, $Y$ could be the Hewitt-realcompactification of $X$).
Let $C_{n}=CL_{Y}(L_{n})$ for all $n$. Let $D=\bigcup_{n\in\omega}C_{n}$. Then $D$ is an $F_{\sigma}$-set in $Y$, so $D$ is realcompact. However, each $C_{n}$ is nowhere dense in $D$, so $D$ is not a Baire space. In fact, since every $F_{\sigma}$ subset of a paracompact space is paracompact, if the space $Y$ is paracompact, then the obtained space $D$ would be paracompact but a countable union of nowhere dense sets.
Let me close with a couple propositions that further show that non-Baire uniform spaces and non-Baire complete uniform spaces are quite common.
$\mathbf{Proposition}$ Suppose that $X$ is a regular space without any isolated points such that for each $x\in X$ there is a collection $\mathcal{U}$ of pairwise disjoint open sets and $x_{U}\in U$ for each $U\in\mathcal{U}$ (i.e. every point is the limit of a discrete set) such that $x\in\overline{\{x_{U}|U\in\mathcal{U}\}}\setminus\{x_{U}|U\in\mathcal{U}\}$. Then for each $x_{0}\in X$ there is a non-Baire subspace containing $x_{0}$.
$\mathbf{Proof}$ We shall construct a sequence of disjoint sets $(C_{n})_{n\in\omega}$ along with open neighborhoods $U_{x}$ of $x$ for each $x\in X$ by induction on $n$.
For $n=0$, let $x_{0}\in X$, let $C_{0}=\{x_{0}\}$ and let $U$ be an open neighborhood of $x_{0}$.
Now assume that $n>0$ and $C_{n-1}$ along with $U_{x}$ for $x\in C_{n-1}$ has been constructed already. Then
for each $x\in C_{n-1}$, let $L_{x}\subseteq U_{x}$ be a subset and let $(U_{z})_{z\in L_{x}}$ be a collection of sets with $\overline{U_{z_{1}}}\cap\overline{U_{z_{2}}}=\emptyset$ and such that $z\in U_{z}\subseteq\overline{U_{z}}\subseteq U_{x}$ for each $z\in L_{x}$. Let $C_{n}=\bigcup_{x\in C_{n-1}}L_{x}$
Since each $x\in C_{n-1}$ is non-isolated in $C_{n-1}\cup C_{n}$ and $U_{x}\cap C_{n-1}=\{x\}$, the set $C_{n-1}$ is nowhere dense in $C_{n-1}\cup C_{n}$. Therefore if $D=\bigcup_{n\in\omega}C_{n}$, then each set $C_{n}$ is is nowhere dense in $D$, so $D$ is not a Baire space. $\mathbf{QED}$.
$\mathbf{Proposition}$ Suppose that $X$ is a normal space without any isolated points such that for each nowhere dense subspace $C\subseteq X$ there is a collection $\mathcal{U}$ of pairwise disjoint open subsets of $X$ along with closed nowhere dense subsets $C_{U}\subseteq U$ such that $C\subseteq\overline{\bigcup\{C_{U}|U\in\mathcal{U}\}}\setminus\bigcup\{C_{U}|U\in\mathcal{U}\}.$ Then for each nowhere dense subspace $C\subseteq X$ there is a $F_{\sigma}$ set $D\subseteq X$ with $C\subseteq D$ and where $D$ is a non-Baire space and $D$ is a $F_{\sigma}$ subset of $X$.
