Decimal form of irrational numbers

Certainly not, since rational numbers are exactly the ones with an eventually periodic expansion. So for example, the number $0.01001100011100001111\ldots$ is irrational.

"Almost every" real number has every digit from $0$ to $9$, in the sense that the set of real numbers that do not has Lebesgue measure $0$.

To show this, it suffices (by countable additivity of the measure) to show that the set of real numbers between 0 and 1 that do not contain every digit has measure zero. In turn it suffices by additivity of the measure to show that the set of real numbers between 0 and 1 with no $9$s has measure $0$. Call this set $S$.

Consider a decimal expansion $0.a_1a_2a_3 \ldots$. The set of real numbers for which $a_1, a_2, a_3, \ldots , a_{k-1} \ne 9$ and $a_k = 9$, call it $E_k$, is measurable and has measure exactly $9^{k-1} \cdot \frac{1}{10^k}$. A real number has the digit $9$ in this decimal expansion if and only if it is not in $E_k$ for any $k$, and the $E_k$ are disjoint. Thus, $$ \mu([0,1] \setminus S) = \mu\left(\bigcup_{k=1}^\infty E_k \right) = \sum_{k=1}^\infty \mu(E_k) = \sum_{k=1}^\infty \frac{9^{k-1} }{10^k} = 1 $$ and $S$ has measure $0$ as desired.

You ask whether the property you describe is true "in general for irrational numbers". That phrase can mean "all irrational numbers", or "most irrational numbers". Because you have accepted Geoffrey's answer of "Certainly not...", I guess you mean "all".

However, to be clear, I want to give treatment to "most" because the answer to that question is "Yes..."

Consider an irrational number like

$$x = 0.1280451740318436570487162...$$

that contains no 9s. We'll call such numbers 9-less. From this single number, many 9-full irrationals can be created simply by inserting 9s in various places.

$x$ is non-terminating. Therefore, there is an infinitude of possible insertion points, which means an infinitude of 9-full irrationals can be generated in this way. (In fact, it can be shown that the set is uncountable.)

Each 9-less irrational generates a distinct infinite set of 9-full irrationals. Therefore, the ratio of 9-less irrationals to 9-full irrationals is zero.

Every irrational number is either 9-less or 9-full, but not both, so almost all irrational numbers are 9-full.

Addendum:

As you will find in the Wikipedia article, "almost all" has several different uses. In this case it means 9-less numbers have Lebesgue measure 0. In other words, they're "sprinkled" among the irrationals, no two "touching". It also means that if you pick an irrational number at random (from a finite interval), it has probability 0 of being 9-less, and probability 1 of being 9-full. This is despite the fact that both sets have the same size, or cardinality. Such is the weirdness of infinities.

Further addendum:

There's another proof of this involving the probability of selecting an infinite sequence of digits randomly and never getting a 9.

This gives a nice opportunity to use cardinality arguments to show that the answer is negative.

Pick two digits $n,k$ such that $\{n,k\}\neq\{0,9\}$. There is a bijection between the numbers in $[0,1]$ whose decimal form includes only $n$ and $k$, and the set of infinite binary sequences.

Therefore the set $\{x\in[0,1]\mid x\text{ has only }n,k\text{ as decimal digits}\}$ is uncountable. So it must include at least one irrational number, and in fact almost the entire set is made of irrational numbers.

The same can be done with three, four, five, six, seven, eight or nine digits. The argument is just the same. Alternatively one can use diagonalization to show that the set is uncountable, it's the same method as you would use otherwise.