No.

A convenient way to see this is via the subadditive (almost additive) thermodynamic formalism. There are natural quasi-Bernoulli equilibrium measures that come from almost additive potentials, and in general these cannot be realised as Gibbs measures for any single continuous (additive) potential.

Outline.

1. Almost additive potentials and their equilibrium states.
   On the full shift X = {0,1}^N, let Φ = {φn}n≥1 be an almost additive potential, i.e. there exists K ≥ 0 such that for all x ∈ X and m,n ∈ N, |φn+m(x) − φn(x) − φm(σn x)| ≤ K, and such that each φn has uniformly bounded variation on n-cylinders (the “Bowen property” for the sequence). For such Φ, there is a well-developed thermodynamic formalism (see e.g. Y. Cao, D.-J. Feng, W. Huang, The thermodynamic formalism for sub-additive potentials, Ergodic Theory Dynam. Systems 28 (2008), 1007–1039). In particular, there exist equilibrium states μΦ which satisfy a Gibbs-type estimate for Φ: C−1 ≤ μΦ[i1…in] / exp(−nP(Φ) + φn(x)) ≤ C for all x ∈ [i1…in] and all n, with C independent of x,n. From this, one immediately gets the quasi-Bernoulli property C′−1 μΦ[ij] ≤ μΦ[i] μΦ[j] ≤ C′ μΦ[ij] with a constant C′ independent of i,j. Thus μΦ is quasi-Bernoulli.

2. Why such a μΦ need not be Gibbs for any single continuous f.
   If μ were a Gibbs measure for some continuous f (in your sense, i.e. with a uniform constant C for all cylinders and lengths), then necessarily f has summable variations: indeed, from C−1 ≤ μ[i1…in]/exp(−nP + Σk=0n−1 f(σk x)) ≤ C one gets for x,y in the same n-cylinder |Σk=0n−1 f(σk x) − Σk=0n−1 f(σk y)| ≤ 2 log C, hence supn Σj=1n varj(f) < ∞ and therefore Σj≥1 varj(f) < ∞. In particular, for such an f and the above μ, we would have that −log μ[i1…in] = nP − Σk=0n−1 f(σk x) + O(1) uniformly on cylinders. Equivalently, the almost additive sequence ψn(x) := −log μ[x0…xn−1] would differ from the additive sequence Σk=0n−1 f(σk x) by a uniformly bounded amount: supx,n |ψn(x) − Σk=0n−1 f(σk x)| < ∞. In other words, ψ is “cohomologous up to a bounded error” to an additive potential.

Now choose Φ for which this is known to be impossible: there exist almost additive (in fact, subadditive) potentials which are not cohomologous to any additive potential in this strong, uniform sense. A standard and very explicit source of such examples comes from linear cocycles: take two 2×2 matrices A0,A1 with strictly positive entries satisfying a cone condition (so that the top Lyapunov exponent behaves with “quasi-multiplicativity”), and define φn(x) := log ||A_{x_{n−1}} … A_{x_0}||. Then {φn} is almost additive (Bowen property holds), and by the cited thermodynamic formalism there is a unique equilibrium state μΦ which satisfies the Gibbs estimate for Φ and hence is quasi-Bernoulli.

However, for a generic choice of such A0,A1, the sequence Φ is not cohomologous (even up to a uniform bounded error) to any additive sequence Σn f∘σk with f continuous; see, for instance,

• I. D. Morris, A necessary and sufficient condition for the subadditive thermodynamic formalism to be additive, Ergodic Theory Dynam. Systems 34 (2014), 1265–1280, which gives precise conditions guaranteeing non-additivity of the subadditive formalism for matrix cocycles. In particular, in these cases there is no continuous f with supx,n |φn(x) − Σk=0n−1 f(σk x)| < ∞.

Putting the two parts together: μΦ is quasi-Bernoulli, but if it were Gibbs for some continuous f in your sense, then the preceding uniform bounded-difference relation would have to hold, contradicting the non-additivity result. Hence μΦ is not Gibbs for any continuous potential.

Remarks.

• The above construction can be made fully explicit by choosing concrete positive matrices A0,A1 (for which the quasi-multiplicativity and non-additivity are known; see, e.g., work of Käenmäki and coauthors on equilibrium states for matrix products). The positivity/cone condition ensures the needed bounded distortion on cylinders and hence the quasi-Bernoulli property of the equilibrium state.

• As noted, if a measure satisfies your Gibbs property for some continuous f with a uniform constant C independent of n, then f automatically has summable variations. So the counterexample above also shows there are quasi-Bernoulli measures whose “log-Jacobian” cannot come from any continuous function with summable variations.

So the answer is no: there exist quasi-Bernoulli measures which are not Gibbs for any continuous potential in the (uniform) Bowen sense.