I don't know a reference to a result. But the statement is simple enough to just give a proof here.

First observe that $f$ is necessarily convex (taking the limit of $f_i(tx + (1-t)y) \leq tf_i(x) + (1-t)f_i(y)$), and non-monotonic. This implies that $f$ attains its infimum [Roc, Section 27]: there exists $x_0$ such that $f(x_0) \leq f(x)$ for all $x\in \mathbb{R}$.

We now prove the claim by contradiction. Assume that $\inf f_i \not\to \inf f$. Then at least one of $\limsup_i \inf_{\mathbb{R}} f_i > \inf f$ or $\liminf_i \inf_{\mathbb{R}} f_i < \inf f$ is true, we call them cases 1 and 2.

Case 1: after taking a subsequence $f_{i'}$ we have that there exists a fixed constant $\epsilon$ such that $f_{i'}(x) > \inf_{\mathbb{R}} f_{i'} > \inf f + \epsilon$. This contradicts $f_{i'}(x_0) \to f(x_0) = \inf f$.

Case 2: after taking a subsequence $f_{i'}$ we have that there exists a fixed constant $\epsilon$ such that $\inf_{\mathbb{R}} f_{i'} < \inf f - \epsilon$. By definition there exists $x_{i'}$ such that $f_{i'}(x_{i'}) \leq \inf f_{i'} + \epsilon / 2$. Now, if $x_{i'}$ accumulates at $x_{\infty}$, then $f(x_{\infty}) \leq \inf f_{i'} + \epsilon / 2 < \inf f$ is a contradiction. So $x_{i'}$ cannot accumulate, so that only finitely many can belong to every closed interval. Without loss of generality we can assume that a subsequence $x_{i''} \to +\infty$. By convexity $f_{i''}|_{[x_0,x_{i''}]} \leq \inf f$ hence pointwise convergence implies that $f |_{[x_0,\infty)} = f(x_0)$. This contradicts non-monotonicity of $f$.

[Roc]: Rockafeller, *Convex Analysis*

Note that I've assumed that non-monotonicity means non-weak-monotonicity. If non-monotonicity allows functions that are weakly, but not strongly, monotonic, then the statement is false: if you require $\inf f_n > -\infty$, you can let $f(x) = 0$ if $x < 0$ and $f(x) = x$ for $x\geq 0$. And define $f_n$ by

$$ f_n(x) = \begin{cases}
-1 & x < -n \\
x / n & -n \leq x < 0 \\
x & 0 \leq x \end{cases} $$