$f$ locally (Lebesgue) integrable function on real line, $g(x):= \lim _{r\to \infty} \frac 1r \int_{x-r}^{x+r} f(t) dt$ exists for every real $x$
The answer is yes.
Let $\int_a^b:=\int_a^b f(t) dt$, with $\int_b^a:=-\int_a^b$ (if $a<b$). We have
\begin{equation}
\int_{x-r}^{x+r}\sim rg(x)\tag{*}
\end{equation}
as $r\to\infty$, where $A\sim B$ is understood as $A-B=o(r)$, and $x,y,\dots$ are any real numbers. Hence, substituting $s:=r-x$, we have
\begin{equation}
\int_{-s}^{2x+s}\sim rg(x)\sim sg(x)
\end{equation}
as $s\to\infty$. Comparing this with
\begin{equation}
\int_{-s}^{s}\sim sg(0),
\end{equation}
we have
\begin{equation}
\int_{s}^{2x+s}\sim s(g(x)-g(0))
\end{equation}
for $s\to\infty$ -- and then similarly for $s\to-\infty$.
Hence,
\begin{equation}
\int_{y-r}^{y+r}-\int_{x-r}^{x+r}=\int_{x+r}^{y+r}-\int_{x-r}^{y-r}
\sim2r(g(\tfrac{y-x}2)-g(0))
\end{equation}
as $r\to\infty$.
Also, by $(*)$,
\begin{equation}
\int_{y-r}^{y+r}-\int_{x-r}^{x+r}
\sim r(g(y)-g(x)).
\end{equation}
So,
\begin{equation}
g(y)-g(x)=2[g(\tfrac{y-x}2)-g(0)] \tag{**}
\end{equation}
for all real $x,y$. Since $g$ is measurable, this implies that it is affine -- see details in the edit below.
Edit: Willie Wong offered a strengthening of the above argument, which yields the Cauchy functional equation, from which the result easily follows.
I would like to show how to use just the functional equation $(**)$ to get the same result. The reasoning is quite similar to that for the Cauchy functional equation; cf. this. Indeed, rewrite $(**)$ as \begin{equation} h(y)-h(x)=2h(\tfrac{y-x}2) \tag{***} \end{equation} for all real $x,y$, where $h:=g-g(0)$. Since $h(0)=0$ and $h(y)-h(x)$ depends on $x$ and $y$ only via $y-x$, it easily follows that $h(x)=h(1)x$ for all rational $x$.
It remains to show that $h$ is continuous. In view of $(***)$, it is enough to verify that $h$ is continuous at $0$. Take any neighborhood $U$ of $0$, and then let $V$ be any neighborhood of $0$ such that $\frac12\,(V-V)\subseteq U$. The union of the sets $h^{-1}(x+V)$ over all rational $x$ is $\mathbb R$. So, for some rational $q$ the set $W:=h^{-1}(q+V)$ is of positive Lebesgue measure. Therefore, $W-W$ is a neighborhood of $0$. By $(***)$, for any $x,y$ in $W$, we have $h(\tfrac{y-x}2)=\frac12\,(h(y)-h(x))\in\frac12\,[(q+V)-(q+V)]=\frac12\,(V-V)\subseteq U$, so that $h(\frac12\,(W-W))\subseteq U$. Since $W-W$ is a neighborhood of $0$, the result follows.