Heuristically false conjectures I was very surprised when I first encountered the Mertens conjecture. Define 
$$ M(n) = \sum_{k=1}^n \mu(k) $$
The Mertens conjecture was that $|M(n)| < \sqrt{n}$ for $n>1$, in contrast to the Riemann Hypothesis, which is equivalent to $M(n) = O(n^{\frac12 + \epsilon})$ .  
The reason I found this conjecture surprising is that it fails heuristically if you assume the Mobius function is randomly $\pm1$ or $0$. The analogue fails with probability $1$ for a random $-1,0,1$ sequence where the nonzero terms have positive density. The law of the iterated logarithm suggests that counterexamples are large but occur with probability 1. So, it doesn't seem surprising that it's false, and that the first counterexamples are uncomfortably large. 
There are many heuristics you can use to conjecture that the digits of $\pi$, the distribution of primes, zeros of $\zeta$ etc. seem random. I believe random matrix theory in physics started when people asked whether the properties of particular high-dimensional matrices were special or just what you would expect of random matrices. Sometimes the right random model isn't obvious, and it's not clear to me when to say that an heuristic is reasonable. 
On the other hand, if you conjecture that all naturally arising transcendentals have simple continued fractions which appear random, then you would be wrong, since $e = [2;1,2,1,1,4,1,1,6,...,1,1,2n,...]$, and a few numbers algebraically related to $e$ have similar simple continued fraction expansions. 
What other plausible conjectures or proven results can be framed as heuristically false according to a reasonable probability model?
 A: CS theory has a slew of these examples.  In particular, take any problem which is known to be in $RP$, but its membership in $P$ is (currently) unknown.
Example: is it possible, using walks consisting of polynomially many steps, to estimate the volume of a convex body?
In the terminology of your question, the answer is 'yes' if you say that random steps are a reasonable model of the steps made by a smart algorithm.  On the other hand, a deterministic method of choosing the steps is unknown.
(PS the reference on this particular problem is "A random polynomial-time algorithm for approximating the volume of convex bodies" by Dyer, Frieze, Kannan.)
A: The Alon-Tarsi Conjecture states that the number of even Latin squares is not equal to the number of odd Latin squares for even $n$.  Although, it can be shown that the gcd of these two numbers grows super-exponentially with $n$ (i.e. these two numbers have many common divisors).  Moreover, it seems that they're asymptotic (using an heuristic argument).
A: Just run across this question, and am surprised that the first example
that came to mind was not mentioned:
Fermat's "Last Theorem" is heuristically true for $n > 3$,
but heuristically false for $n=3$ which is one of the easier
cases to prove.
if $0 < x \leq y < z \in (M/2,M]$ then $|x^n + y^n - z^n| < M^n$.
There are about $cM^3$ candidates $(x,y,z)$ in this range
for some $c>0$ (as it happens $c=7/48$), producing values of
$\Delta := x^n+y^n-z^n$ spread out on
the interval $(-M^n,M^n)$ according to some fixed distribution
$w_n(r) dr$ on $(-1,1)$ scaled by a factor $M^n$ (i.e.,
for any $r_1,r_2$ with $-1 \leq r_1 \leq r_2 \leq 1$
the fraction of $\Delta$ values in $(r_1 M^n, r_2 M^n)$
approaches $\int_{r_1}^{r_2} w_n(r) dr$ as $M \rightarrow \infty$).
This suggests that any given value of $\Delta$, such as $0$,
will arise about $c w_n(0) M^{3-n}$ times.  Taking $M=2^k=2,4,8,16,\ldots$
and summing over positive integers $k$ yields a rapidly divergent sum
for $n<3$, a barely divergent one for $n=3$, and a rapidly convergent
sum for $n>3$.
Specifically, we expect the number of solutions of $x^n+y^n=z^n$
with $z \leq M$ to grow as $M^{3-n}$ for $n<3$ (which is true and easy),
to grow as $\log M$ for $n=3$ (which is false), and to be finite for $n>3$
(which is true for relatively prime $x,y,z$ and very hard to prove [Faltings]).
More generally, this kind of analysis suggests that for $m \geq 3$
the equation $x_1^n + x_2^n + \cdots + x_{m-1}^n = x_m^n$
should have lots of solutions for $n<m$,
infinitely but only logarithmically many for $n=m$,
and finitely many for $n>m$.  In particular, Euler's conjecture
that there are no solutions for $m=n$ is heuristically false for all $m$.
So far it is known to be false only for $m=4$ and $m=5$.
Generalization in a different direction suggests that any cubic
plane curve $C: P(x,y,z)=0$ should have infinitely many rational points.
This is known to be true for some $C$ and false for others;
and when true the number of points of height up to $M$ grows as
$\log^{r/2} M$ for some integer $r>0$ (the rank of the elliptic curve),
which may equal $2$ as the heuristic predicts but doesn't have to.
The rank is predicted by the celebrated conjecture of Birch and
Swinnerton-Dyer, which in effect refines the heuristic by accounting
for the distribution of values of $P(x,y,z)$ not just
"at the archimedean place" (how big is it?) but also "at finite places"
(is $P$ a multiple of $p^e$?).
The same refinement is available for equations in more variables,
such as Euler's generalization of the Fermat equation;
but this does not change the conclusion (except for equations such as
$x_1^4 + 3 x_2^4 + 9 x_3^4 = 27 x_4^4$,
which have no solutions at all for congruence reasons),
though in the borderline case $m=n$ the expected power of $\log M$ might rise.
Warning: there are subtler obstructions that may prevent a surface from
having rational points even when the heuristic leads us to expect
plentiful solutions and there are no congruence conditions that
contradict this guess.  An example is the Cassels-Guy cubic
$5x^3 + 9y^3 + 10z^3 + 12w^3 = 0$, with no nonzero rational solutions
$(x,y,z,w)$:

Cassels, J.W.S, and Guy, M.J.T.:
  On the Hasse principle for cubic surfaces,
  Mathematika 13 (1966), 111--120.

A: This is quite elementary, but surprised me when I first saw it, and I still think it's remarkable. 
The number of pairs of integers $(x, y)$ such that $x^2 + y^2 \leq n$ is asymptotically $\pi n$, since they are the lattice points inside a circle of radius $\sqrt{n}$. Therefore the average number of ways of writing a positive integer as a sum of two squares is $\pi$. Or $\pi/8$ if we regard solutions as the same when they differ only in signs or the order of the terms.  
One would therefore expect a positive proportion of the natural numbers to have a representation as a sum of two squares. Not a $\pi/8$-fraction, since some integers have several representations, but some slightly smaller positive density, since identities like $4^2 + 7^2 = 1^2 + 8^2$ look pretty much like random coincidences.
But actually almost no numbers are sums of two squares. Whenever the prime factorization of $n$ contains some prime $p\equiv 3$ (mod 4) to an odd power, $n$ cannot be a sum of two squares, as is easily seen by considering the equation modulo powers of $p$. And by Dirichlet's theorem, almost all numbers have some such prime to power 1 in their factorization.
A: I think this example fits, in 1985 H. Maier disproved a very reasonable conjecture on the distribution of prime numbers in short intervals. The probabilistic approach had been thoroughly examined by Harald Cramer. Nice paper by Andrew Granville including this episode in  (mathematical) detail, page 23 (or 13 out of 18 in the pdf): 
www.dms.umontreal.ca/~andrew/PDF/cramer.pdf
