I am a bit confused. Your example with $m = 2$ does not support your claim that the mean of the sequence exceeds $m/2 = 1$, since $4/7 < 1$.
Furthermore, I should point out that under the assumption that the sequence is uniformly distributed, the mean would be $(m-1)/2$, not $m/2$, because no element of the sequence can attain $m$; all values are in $\{0, 1, \ldots, m-1\}$. So under this modification, yes, your counterexample does work.
Moreover, there are many less trivial counterexamples that are easily found, in both directions: that is to say, the mean can be quite a bit lower than $(m-1)/2$, as well as larger: if $m = 67$, then the period is only $P(m) = 33$, and the sequence consists of $$\{1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 21, 62, 55, 9, 4, 59, 1, 5, 64, 65, 3, 0, 65, 1, 1, 66, 0, 1, 0, 0 \}.$$ Its mean is $\frac{670}{33} \approx 20.303$ which is significantly smaller than $(67-1)/2 = 33$. In the other direction, $m = 159$ has a period of $P(m) = 936$ (its length is too much to include here in its entirety), with mean $\frac{12455}{156} \approx 79.8397$, exceeding $(159-1)/2 = 79$. On the basis of these small-$m$ examples, we have no reason to expect that this generator should be unbiased, let alone uniformly distributed on $\{0, \ldots, m-1\}$.
What would be interesting to try to answer, however, is whether the behavior of this generator is asymptotically unbiased for large $m$, in the sense that $$\limsup_{m} \left|\bar F_{m} - \frac{m - 1}{2}\right| = 0$$ where $\bar F_{m}$ represents the mean of the lagged Fibonacci generator (for your given choice of parameters) over its period? I am going to guess that it is not, based on the results of the following plot of $(m-1)/2 - \bar F_m$ for $m = 2, \ldots, 400$: 
