Does special relativity imply that time dilation is affected by an orientation of clocks?
Answer to the title question: absolutely not.
Experimental Touchstone
Before we explain in detail, let's begin by noticing that the Michelson-Morley interferometric experiment explicitly tests if orientation affects the clocking behavior of a there-and-back light path. And quite famously the answer is "no". This must be true for any inertial observer.1
So why do all the introductory materials use a transverse clock?
It's actually a good question and the answer (at least beyond "Well, that's what Einstein did!"), requires taking a close look at the way the explanation would work with a longitudinal clock.
What is Going on, Then?
The short version is easy: because the longitudinal light clock is affected by length contractions as well as time dilation.2 And then it follows that from a didactic point of view you want to develop one of the rules (time dilation or length contraction) first, and address the second one separately rather than trying to deal with them at the same time. That makes the transverse clock preferable for intorducing relativity.
To show this the long way we'll imagine two basically identical light-reflection clocks $\mathbb{c}$ and $\mathbb{C}$, where $\mathbb{c}$ is the traditional transverse clock and $\mathbb{C}$ is aligned longitudinally.3 In their rest frame $S$, each clock has length $l = L$, and consequently identical periods $p = 2l/c$ and $P = 2L/c$. We then consider the behavior of the clocks as observed in frame $S'$ moving at speed $-v$ along the length of $\mathbb{C}$ with respect to $S$.
Transverse case
The analysis of the period of $p'$ of the transverse clock is the traditional one: the time required to complete the trip (out and back) is \begin{align} p' &= \frac{\sqrt{(2l)^2 + (vp')^2}}{c}\\ &= \sqrt{\left(\frac{2l}{c}\right)^2 + \left(\beta p'\right)^2} \\ &= p \sqrt{1 + \left( \beta \frac{p'}{p}\right)^2} \;, \end{align} so that \begin{align} \left(\frac{p'}{p} \right)^2 &= 1 + \left(\beta \frac{p'}{p}\right)^2\\ \frac{p'}{p} &= \left(1 - \beta^2 \right)^{-1/2} \\ &= \gamma\;. \end{align}
Longitudinal case
To find the period $P'$ of the longitudinal clock we have to do a bit more figuring. The elapsed time $T_f$ for the forward going half of the journey is $$ T_f' = \frac{L' + v T_f'}{c} \;,$$ and for the backward going half of the journey the time $T_b$ required is $$ T_b' = \frac{L' - v T_b'}{c} \;.$$ After a little figuring we get the period as \begin{align} P' &= \frac{L'}{c(1 - \beta)} + \frac{L'}{c(1 + \beta)}\\ &= \frac{2L'}{c(1 - \beta^2)} . \end{align} Now, if $L' = L$ this would lead to $$ \frac{P'}{P} = \left( 1 - \beta^2 \right)^{-1} = \gamma^2 \;, \tag{wrong!}$$ meaning the clocks would not agree, but as we said before Michelson-Morley style of experiments rule that out, so $L'$ must not be the same as $L$. To get the agreement we must have it is required that $$ \frac{L'}{L} = \left(1 - \beta^2 \right)^{1/2} \;,$$ the usual expression for length contraction.
Better Way
All that work is, quite frankly, nasty, and I would recommend a geometry first approach as a better alternative to Einstein's version. Get Takeuchi's book, it's worth the money.
1 Because it tells us that two clocks set with their emit/receive ends coincident that beat in time with one another will still beat in time with one another when you swing them around. It doesn't mean that all observer will agree on the frequency of the clocks, just that the two clocks agree.
2 That's what Lorentz-Fitzgerald contraction is all about after-all: fixing up the classical theory to match the Michelson-Morley results.
3 We'll continue to use lower case for quantities related to the transverse clock and capitals for quantities related to the longitudinal clock throughout.