The maps $(\sigma _n)$ are asymptotically constant, in the sense that there is a sequence $(y_n)$ in $Y$ for which $d(\sigma _n , y_n )\rightarrow 0$ in measure, in each of the following cases:
$\bullet$ $(Y,d)$ is compact,
$\bullet$ $(Y,d)$ is an ultrametric space,
$\bullet$ $Y$ is a locally compact second countable group with finite asymptotic dimension (and $d$ is a compatible proper left-invariant metric on $Y$).
The assumption that the convergence $d(\gamma \cdot \sigma _n , \sigma _n )\rightarrow 0$ is pointwise a.e. for each $\gamma \in \Gamma$ (where $(\gamma\cdot \sigma _n )(x)=\sigma _n (\gamma ^{-1}x)$) can be relaxed to assuming that this convergence is in measure for each $\gamma \in \Gamma$.
$\bullet$ Suppose $(Y,d)$ is compact. I'll sketch an argument for this case using ultraproducts (a different argument can be obtained by seeing this as a special case of Lemma 2.4 of this paper by Adrian Ioana and myself). Note that it suffices to prove the conclusion along a subsequence of $(\sigma _n)$. By assumption $\int _X d(\sigma _n (g^{-1}x), \sigma _n(x)) \, d\mu \rightarrow 0$ for all $g\in \Gamma$. Let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{N}$. Then the strong ergodicity of the action $\Gamma \curvearrowright (X,\mu )$ is equivalent to ergodicity of the ultrapower action $\Gamma \curvearrowright (X_{\mathcal{U}},\mu _{\mathcal{U}})$. Since $Y$ is compact we can define a map $\sigma :X_{\mathcal{U}}\rightarrow Y$ by $\sigma ([x_n])=\lim _{n\rightarrow \mathcal{U}}\sigma _n (x_n)$, which is measurable (here $[x_n]\in X_{\mathcal{U}}$ denotes the equivalence class of a sequence $(x_n)\in X^{\mathbb{N}}$), and for each $g\in \Gamma$ we have
$\int _{X_{\mathcal{U}}}d(\sigma (g[x_n]),\sigma ([x_n]))\, d\mu _{\mathcal{U}} = \lim _{n\rightarrow \mathcal{U}}\int _X d (\sigma _n(gx), \sigma _n (x))\, d\mu = 0$.
Therefore, by ergodicity there is some $y\in Y$ such that $\lim _{n\rightarrow \mathcal{U}}\sigma _n (x_n)= y$ for a.e. $[x_n]\in X_{\mathcal{U}}$. Thus, $\lim _{n\rightarrow \mathcal{U}}\int _X d (\sigma _n (x),y ) \, d\mu = \int _{X_{\mathcal{U}}} d(\sigma ([x_n]),y) \, d\mu _{\mathcal{U}} = 0$, which finishes the proof. $\square$
Note that this gives a proof of the well known fact that strong ergodicity of $\Gamma \curvearrowright (X,\mu )$ implies that if $f_n \in L^{\infty}(X,\mu )$ satisfies $\sup _n \| f_n \| _{\infty}< \infty$ and $\| g\cdot f_n - f_n \| _1\rightarrow 0$ for all $g\in \Gamma$, then $\| f_n - \int f_n \, d\mu \| _1\rightarrow 0$. (Here, $(g\cdot f_n)(x)= f_n(g^{-1}x)$.)
$\bullet$ Suppose $(Y,d)$ is a (separable) ultrametric space. The argument in this case is essentially the same as Lemma 2.5 from this paper of Adrian Ioana. Given $\epsilon >0$, it is enough to find a finite $F\subseteq \Gamma$ and $\delta >0$ such that if $\sigma :X\rightarrow Y$ is a measurable map satisfying $\mu ( \{ x : d(\sigma (g^{-1}x),\sigma (x)) <\delta \} )>1-\delta$ for all $g\in F$, then there is some $y\in Y$ with $\mu ( \{ x : d(\sigma (x),y)<\epsilon \} )>1-\epsilon$. We may assume that $\epsilon < 1/3$.
By strong ergodicity we can find a finite $F\subseteq \Gamma$ and $\delta >0$ such that if $A\subseteq X$ satifies $\mu (gA\triangle A )<\delta$ for all $g\in F$ then either $\mu (A)<\epsilon$ or $\mu (A)>1-\epsilon$. We may assume that $\delta <\epsilon$. Suppose that $\sigma : X\rightarrow Y$ is as above, using this $\delta$.
Let $\mathcal{B}_{\epsilon}$ denote the collection of all $\epsilon$-balls in $Y$, so that $\mathcal{B}_{\epsilon}$ is a countable partition of $Y$ with $d(A,B)\geq \epsilon$ for distinct $A,B\in \mathcal{B}_{\epsilon}$. Let $\tau (x) \in \mathcal{B}_{\epsilon}$ be the unique element containing $\sigma (x)$. Then for each $g\in F$, the assumption on $\sigma$ implies that $\mu ( \{ x : \tau (g^{-1}x)=\tau (x) \} ) > 1-\delta$, and hence for every subset $\mathcal{D}$ of $\mathcal{B}_{\epsilon}$ we have $\max _{g\in F} \mu (g\tau ^{-1}(\mathcal{D})\triangle \tau ^{-1}(\mathcal{D}) )<\delta$. So, by our choice of $\delta$, the measure $\tau _*\mu$ on $\mathcal{B}_{\epsilon}$ takes values in the interval $[0,\epsilon )\cup (1-\epsilon ,1]$, and hence there is some $B\in \mathcal{B}_{\epsilon}$ with $\mu (\tau ^{-1}(B))>1-\epsilon$. Fixing some $y\in B$, this means that $\mu ( \{ x : d(\sigma (x),y)<\epsilon \} )>1-\epsilon$, as was to be shown. $\square$.
$\bullet$ The last case mentioned above is a consequence of the following Lemma: Let $\Gamma\curvearrowright (X,\mu )$ be strongly ergodic, let $(Y,d)$ be a separable metric space, and let $\sigma _n : X\rightarrow Y$ be a sequence of measurable maps with $d(g\cdot \sigma _n , \sigma _n)\rightarrow 0$ in measure for all $g\in \Gamma$. Suppose that $\mathcal{C}$ is a countable cover of $Y$ that can be written as a finite union $\mathcal{C}=\bigcup _{i=0}^m\mathcal{C}_i$ where for each $i$ the collection $\mathcal{C}_i$ is uniformly disjoint, i.e., there is some $r>0$ such that $d(A,B)\geq r$ for all distinct $A,B\in \mathcal{C}_i$. Then there exists a sequence $(D_n)$ in $\mathcal{C}$ such that $d(\sigma _n , D_n )\rightarrow 0$ in measure.
For the proof of this lemma, after replacing $d$ by the uniformly equivalent metric $d/(1+d)$ if necessary, we may assume that $d\leq 1$. For each $i$, letting $C_i=\bigcup \mathcal{C}_i$, we have $\| d(g\cdot \sigma _n , C_i ) - d(\sigma _n , C_i ) \| _1\leq \| d(g\cdot \sigma _n ,\sigma _n ) \| _1 \rightarrow 0$ for all $g\in G$. Thus (by the note after the compact case) $\| d(\sigma _n, C_i ) - \widetilde{d}(\sigma _n, C_i ) \| _1 \rightarrow 0$ for all $i\leq m$, where $\widetilde{d}(\sigma _n, C_i ) = \int d(\sigma _n (x),C_i )\, d\mu$. Since $C_0,\dots , C_m$ cover $Y$, this implies that for every $\epsilon >0$ there are at most finitely many $n$ for which $\min _{i\leq m} \widetilde{d}(\sigma _n , C_i) \geq \epsilon$. Thus, there must be some sequence $(i_n)$ such that $\widetilde{d}(\sigma _n, C_{i_n} ) \rightarrow 0$. By perturbing the $\sigma _n$ an asymptotically negligible amount, we may assume that $\sigma _n(x) \in C_{i_n}$ for all $x\in X$ and all $n$. The rest of the proof is exactly as in the ultrametric case (with $\mathcal{C}$ here playing the role of $\mathcal{B}_{\epsilon}$ there), using that each $\mathcal{C}_i$ is uniformly disjoint. $\square$
This applies to the case of a proper metric space $(Y,d)$ of finite asymptotic dimension whose isometry group acts transitively: the finite asymptotic dimension hypothesis is only used once to obtain a single countable cover $\mathcal{C}$ of $Y$ by sets of uniformly bounded diameter, which can be expressed as a finite union $\mathcal{C}=\bigcup _{i=0}^m \mathcal{C}_i$ where each $\mathcal{C}_i$ is uniformly disjoint. By the above lemma, given a sequence $(\sigma _n)$ under consideration, we can find $D_n \in \mathcal{C}$ with $d(\sigma _n , D_n ) \rightarrow 0$ in measure, so by making an asymptotically negligible change we can assume that $\sigma _n (x)\in D_n$ for all $n$. Since the isometry group acts transitively and the $D_n$ are uniformly bounded, we can isometrically translate all the $D_n$ into a single compact ball, and apply the compact case (then translate back) to finish the proof. $\square$
As can be seen from the proof, the assumption of finite asymptotic dimension appears to be overkill. I don't know what happens with general complete separable metric spaces. Perhaps there could be a counterexample using some kind of measure concentration argument, but at the moment I don't have one.
