QPSK and OQPSK Modulation
This question seems to be based on several misconceptions.

In a QPSK modulator operating at a rate of $N$ baud, two bits enter the
modulator during each $T = N^{-1}$ second interval.  If the bits are
entering on a single wire (at a rate of $2N$ bits per second, the
QPSK modulator first converts this serial input
bit stream at $2N$ bits per second into two parallel bit streams at
$N$ bits per second. For example, alternate bits are steered to
alternate streams so that, say, all the odd-numbered bits
are in one stream and all the even-numbered bits are in the 
other stream.

In some cases, the bits entering the QPSK modulator
are already formatted into two parallel bit streams at $N$ bits per
second. One important and commonly observed arrangement of this type
is when a rate-$\frac12$ convolutional encoder precedes the QPSK modulator,
so that information bits enter the encoder serially
at rate $N$ bits per second, and the convolutional encoder spits out two 
parallel bit streams at $N$ bits per second.  However, regardless of
encoding or not, the bits that the QPSK
modulator uses are in two parallel streams that are synchronous
(that is, the bit transitions occur at the same times (once every
$T$ seconds) and in each
bit stream, the bit occupies the entire $T$ second interval.

As the OP's question says, the bits are modulated onto cosine and
sine carriers.  As discussed in more detail here, if the bits in the two streams are denoted by $b_I$ and
$b_Q$, then the corresponding signals, say, over the interval
$[0,T)$ can be taken to be 
$$(-1)^{b_I}\cos(2\pi f_c t) ~~ \text{and}~~ -(-1)^{b_Q}\sin(2\pi f_c t),
0 \leq t &lt; T. \tag{1}$$
(Ignore the negative sign on the $\sin$, it is there to simplify
notation in more detailed analyses).  The QPSK signal is the sum
$$(-1)^{b_I}\cos(2\pi f_c t) -(-1)^{b_Q}\sin(2\pi f_c t),
0 \leq t \leq T. \tag{2}$$
It is not the case, as the OP seems to think that the two
signals are separated into even and odd bits, so that
$(-1)^{b_I}\cos(2\pi f_c t)$ lasts from $t=0$ to $t=T/2$ and
$-(-1)^{b_Q}\sin(2\pi f_c t)$ from $t=T/2$ to $t=T$: the
cosine and sine signals last for the full $T$-second duration
and the bits $b_I$ and $b_Q$ both affect the signal phase for all
$T$ seconds.

Offset QPSK also converts the serial bit stream into
two parallel bit streams in which the bit intervals are
of duration $T$ on each wire, but the bit transitions
are not synchronous but occur with an offset of
$T/2$ seconds.  Thus, $b_I$ modulates the cosine carrier
to produce 
$$(-1)^{b_I}\cos(2\pi f_c t), 0 \leq t &lt; T$$
but $b_Q$ modulates the sine carrier to produce
$$-(-1)^{b_Q}\sin(2\pi f_c t), \frac{T}{2} \leq t &lt; \frac{3T}{2}.$$
The modulator output is still the sum of the two sinusoids,
but the sinusoids change phase at different times. In fact,
the signal can be deemed to be a $2N$ baud signal since the
phase can change once every $T/2 = (2N)^{-1}$ seconds.
Each bit still affects the phase for $T$ seconds, but the
bit transitions are staggered.  In the receiver, integrations
(or matched filterings)
are carried out over (staggered) $T$-second intervals in the
two branches of the receiver, and the decision devices
spit out one bit every $T$ seconds but at staggered times.
It is easy to multiplex these decisions onto a single wire
that thus carries a bit stream at $2N$ bits per second,
just as the input to the QPSK modulator (if anyone still
remembers that far back in this answer) is a bit stream
on a single wire at $2N$ bits per second.