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I believe the Lemma you propose is true, via a relatively straightforward adaptation of the proof given for the additive case in Giles Atkinson, Recurrence of co-cycles and random walks, J. Lond. Math. Soc. (2) 13 (1976), 486-488. (I assume this is the result you were referencing.) This may already be known and written somewhere -- I can't speak to that. In case it's not, here's a proof.

Proof of Lemma. By the subadditive ergodic theorem, there exists a measurable function $f\colon X\to \mathbb{R}\cup\{-\infty\}$ such that $\lim \frac 1n f_n(x) = f(x)$ for $\mu$-a.e. $x\in X$ and $\lim \frac 1n \int f_n\,d\mu = \int f\,d\mu$. So our task is to show that $\int f\,d\mu < 0$.

To this end, given $x\in X$, let $M_x = \{n \mid f_n(x) \geq -1 \}$, and observe that $M_x$ is finite $\mu$-a.e. Thus writing $A_n = \{x\in X \mid \# M_x < n \}$, we see that there exists $N$ such that $\mu(A_N) > \frac 12$.

Furthermore, given $x\in X$, write $L_x = \{ k \mid T^k(x) \in A_N \}$. Since $\mu(A_N) > \frac 12$, we see that $\mu$-a.e. $x$ has $$(*)\qquad\qquad\qquad\# L_x \cap [1,n] \geq \frac n2\qquad\qquad\qquad\qquad\qquad\quad$$ for all sufficiently large $n$. We fix such an $x$ and show that $f(x) < 0$.

Let $k_0$ be the smallest element of $L_x$. We define $k_i\in L_x$ recursively with the property that $$f_{k_i}(x) < f_{k_0}(x)-i,$$ as follows. Let $J_i$ be the $N$ smallest elements of $L_x \cap (k_i,\infty)$. Because $T^{k_i}(x)\in A_N$, there exists $k_{i+1}\in J_i$ such that $k_{i+1} - k_i \notin M_{T^{k_i}(x)}$. In particular, we have $$f_{k_{i+1} - k_i}(T^{k_i}(x)) < -1.$$ Now subadditivity gives $$f_{k_{i+1}}(x) \leq f_{k_i}(x) + f_{k_{i+1} - k_i}(T^{k_i}(x)) < f_{k_0(x)} - i-1.$$ The next observation to make is that by $(*)$, we have $k_i \leq 2Ni$ for all sufficiently large $i$. Thus we have $$\frac 1{k_i} f_{k_i}(x) \leq \frac 1{2Ni}(f_{k_0}(x) - i),$$ and sending $i\to\infty$, we obtain $f(x) \leq -\frac 1{2N}$. Since this holds for $\mu$-a.e. $x$, we have $\int f\,d\mu \leq -\frac 1{2N} < 0$, which completes the proof.

Note. It's quite important in this proof that we're going in the dealing with negative direction; values of $f_n$; the proof would fail if we tried to show that $f_n(x)\to+\infty$ for $\mu$-a.e. $x$ implies that $\int f\,d\mu > 0$. I'm not sure if the result is true in this case.

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I believe the Lemma you propose is true, via a relatively straightforward adaptation of the proof given for the additive case in Giles Atkinson, Recurrence of co-cycles and random walks, J. Lond. Math. Soc. (2) 13 (1976), 486-488. (I assume this is the result you were referencing.) This may already be known and written somewhere -- I can't speak to that. In case it's not, here's a proof.

Proof of Lemma. By the subadditive ergodic theorem, there exists a measurable function $f\colon X\to \mathbb{R}\cup\{-\infty\}$ such that $\lim \frac 1n f_n(x) = f(x)$ for $\mu$-a.e. $x\in X$ and $\lim \frac 1n \int f_n\,d\mu = \int f\,d\mu$. So our task is to show that $\int f\,d\mu < 0$.

To this end, given $x\in X$, let $M_x = \{n \mid f_n(x) \geq -1 \}$, and observe that $M_x$ is finite $\mu$-a.e. Thus writing $A_n = \{x\in X \mid \# M_x < n \}$, we see that there exists $N$ such that $\mu(A_N) > \frac 12$.

Furthermore, given $x\in X$, write $L_x = \{ k \mid T^k(x) \in A_N \}$. Since $\mu(A_N) > \frac 12$, we see that $\mu$-a.e. $x$ has $$(*)\qquad\qquad\qquad\# L_x \cap [1,n] \geq \frac n2\qquad\qquad\qquad\qquad\qquad\quad$$ for all sufficiently large $n$. We fix such an $x$ and show that $f(x) < 0$.

Let $k_0$ be the smallest element of $L_x$. We define $k_i\in L_x$ recursively with the property that $$f_{k_i}(x) < f_{k_0}(x)-i,$$ as follows. Let $J_i$ be the $N$ smallest elements of $L_x \cap (k_i,\infty)$. Because $T^{k_i}(x)\in A_N$, there exists $k_{i+1}\in J_i$ such that $k_{i+1} - k_i \notin M_{T^{k_i}(x)}$. In particular, we have $$f_{k_{i+1} - k_i}(T^{k_i}(x)) < -1.$$ Now subadditivity gives $$f_{k_{i+1}}(x) \leq f_{k_i}(x) + f_{k_{i+1} - k_i}(T^{k_i}(x)) < f_{k_0(x)} - i-1.$$ The next observation to make is that by $(*)$, we have $k_i \leq 2Ni$ for all sufficiently large $i$. Thus we have $$\frac 1{k_i} f_{k_i}(x) \leq \frac 1{2Ni}(f_{k_0}(x) - i),$$ and sending $i\to\infty$, we obtain $f(x) \leq -\frac 1{2N}$. Since this holds for $\mu$-a.e. $x$, we have $\int f\,d\mu \leq -\frac 1{2N} < 0$, which completes the proof.

Note. It's quite important in this proof that we're going in the negative direction; the proof would fail if we tried to show that $f_n(x)\to+\infty$ for $\mu$-a.e. $x$ implies that $\int f\,d\mu > 0$. I'm not sure if the result is true in this case.