Yes, such a map $\alpha$ is automatically multiplicative and thus defines an action of $\widehat{\mathbb{G}}$ on $M$.
As in the question, denote by $\alpha_\gamma : M \to M \otimes B(H_\gamma)$ the components of $\alpha$, for any irreducible unitary representation $\gamma$ of $\mathbb{G}$. Fix an irreducible representation $\gamma$. It suffices to prove that $\alpha_\gamma$ is multiplicative.
Since $\alpha$ is unital completely isometric, $\alpha$ is also completely positive. Thus, all $\alpha_\gamma$ are unital completely positive (ucp).
We first prove that $\alpha_\varepsilon(x) =x$ for all $x \in M$. By the coaction property, $\alpha_\gamma \circ \alpha_\varepsilon = \alpha_\gamma$ for all $\gamma$. So, if $\alpha_\varepsilon(x)=0$, it follows that $\alpha(x)=0$ and thus $x=0$ because $\alpha$ is supposed to be isometric. Since $\alpha_\varepsilon(\alpha_\varepsilon(x)-x) = 0$ for all $x \in M$, it follows that $\alpha_\varepsilon(x) =x$ for all $x \in M$.
Let $\rho$ be the contragredient of $\gamma$ and choose morphisms $t \in \operatorname{Mor}(\varepsilon,\rho \otimes \gamma)$ and $s \in \operatorname{Mor}(\varepsilon,\gamma \otimes \rho)$ such that $t^* t = 1$ and $(s^* \otimes 1)(1 \otimes t) = 1$. Define the ucp map
$$\theta : M \otimes B(H_\gamma) \to M : \theta(x) = (1 \otimes t^*)(\alpha_\rho \otimes \text{id})(x) (1 \otimes t) \; .$$
By the coaction property and the fact that $\alpha_\varepsilon = \text{id}$ proven above, $\theta(\alpha_\gamma(x)) = x$ for all $x \in M$. Fix a unitary $u \in \mathcal{U}(M)$. Since $\theta(\alpha_\gamma(u)) = u$ is a unitary and $\|\alpha_\gamma(u)\| \leq 1$, we find that $\alpha_\gamma(u)$ belongs to the multiplicative domain of $\theta$. We have that $\alpha_\gamma(u)^* \alpha_\gamma(u) \leq \alpha_\gamma(u^*u) = 1$. Applying $\theta$ and using that $\alpha_\gamma(u)$ belongs to the multiplicative domain of $\theta$, we find that
$\theta(1-\alpha_\gamma(u)^* \alpha_\gamma(u)) = 0$. Below I will prove that $\theta$ is faithful. So, we conclude that $\alpha_\gamma(u)^* \alpha_\gamma(u) = 1$ for every unitary $u \in \mathcal{U}(M)$. This implies that $\alpha_\gamma$ is multiplicative.
It remains to prove that $\theta$ is faithful. Assume that $x \in M \otimes B(H_\gamma)$ such that $\theta(x^* x) = 0$. Then, $(\alpha_\rho \otimes \text{id})(x)(1 \otimes t) = 0$. Apply $\alpha_\gamma \otimes \text{id} \otimes \text{id}$ to conclude that
$$(1 \otimes s^* \otimes 1) ((\alpha_\gamma \otimes \text{id})\alpha_\rho \otimes \text{id})(x) (1 \otimes 1 \otimes t) = 0 \; .$$
Using the coaction property of $\alpha$ and the fact that $\alpha_\varepsilon = \text{id}$ as proven above, the left hand side of the above expression equals
$$x (1 \otimes s^* \otimes 1)(1 \otimes 1 \otimes t) = x \; .$$
So $x = 0$ and the faithfulness of $\theta$ is proven.
In the general, not necessarily $\sigma$-compact setting, one has to interpret $L^\infty(G,\lambda)$ as the von Neumann algebra of locally measurable functions that are bounded outside a locally null set. This is very well explained in Section 2.3 of Folland's "A course in abstract harmonic analysis".
Also note that as explained in Proposition 2.4 of that same book, every locally compact group $G$ admits a $\sigma$-compact subgroup $G_0 \subset G$ that is open. So measure theoretically, we are always in the quite tame situation of a (possibly uncountable) disjoint union of copies of $G_0$.
Best Answer
The approach I had in mind is the following. For a reference, see Section 4, Chapter IX of Takesaki 2. The extended positive part $\widehat{M_+}$ is by definition the space of positive-homogeneous, additive, lower semi-continuous maps $M_*^+\rightarrow [0,\infty]$. Given $x\in (M\bar\otimes M)_+$ define $(\iota\otimes\varphi)(x) \in \widehat{M_+}$ to be the map $$ \omega \mapsto \varphi\big( (\omega\otimes\iota)(x) \big). $$ For $\omega\in M_*^+$, as $x$ is positive, also $(\omega\otimes\iota)(x)$ is positive, and so we obtain a well-defined member of $[0,\infty]$. Clearly $(\iota\otimes\varphi)(x)$ is positive-homogeneous and additive. If $(\omega_i)$ is a net in $M_*^+$ converging to $\omega$ in norm, then $(\omega_i\otimes\iota)(x) \rightarrow (\omega\otimes\iota)(x)$ $\sigma$-weakly. As $\varphi$ is $\sigma$-weakly lower semi-continuous (see Theorem 1.11 in Chapter VII of Takesaki) it follows that $\lim_i (\iota\otimes\varphi)(x)(\omega_i) \geq (\iota\otimes\varphi)(x)(\omega)$. Thus $(\iota\otimes\varphi)(x)$ is lower semi-continuous.
Clearly the map $\iota\otimes\varphi$ is positive-homogeneous and additive. Given $a\in M, \omega\in M_*^+$, $$ (\iota\otimes\varphi)((a\otimes 1)^*x(a\otimes 1))(\omega) = \varphi\big( (a \omega a^* \otimes\iota)(x) \big) = (\iota\otimes\varphi)(x)(a \omega a^*) = \big( a^* (\iota\otimes\varphi)(x) a \big)(\omega). $$ Thus $(\iota\otimes\varphi)((a\otimes 1)^*x(a\otimes 1)) = a^* (\iota\otimes\varphi)(x) a$.
Finally, if $(x_i)$ increases to $x$ in $(M\bar\otimes M)_+$ then for each $\omega\in M_*^+$ we have that $(\omega\otimes\iota)(x_i)$ increases to $(\omega\otimes\iota)(x)$ so by normality of $\varphi$, it follows that $(\iota\otimes\varphi)(x_i)(\omega)$ increases to $(\iota\otimes\varphi)(x)(\omega)$. Hence $(\iota\otimes\varphi)(x_i)$ increases to $(\iota\otimes\varphi)(x)$. So $\iota\otimes\varphi$ is normal. So $\iota\otimes\varphi$ is an operator-valued weight.